Applied Optimization - Online Tutor, Practice Problems & Exam Prep

Intro to Applied Optimization: Maximizing Area

In earlier videos, we've worked a lot with the maximum and minimum values of a function. We learned multiple different ways to find these values. It turns out that we can keep applying those same methods when faced with this specific type of word problem, referred to as an applied optimization problem. Any time we encounter a word problem in calculus, it may be a bit intimidating. But I'm going to walk through this problem with you step by step, and we're going to use a lot of what we've already learned.

So let's get into this example. Here, we're told that we have 200 feet of fencing to construct a rectangular fence, and we're asked to determine the dimensions that will create the maximum area. We're told here that one side is actually formed by a rock wall and does not need fencing at all. We're specifically asked to find the maximum area. In other words, we're asked to optimize the real-world value of the area.

How exactly can we do that? Well, we can express the area as a function based on the constraints that are given to us in the problem. Here, we're told that we only have 200 feet to construct this maximum area. With the area expressed as a function, we can then find its maximum using the methods that we already know. So let's get into step 1 here and start solving this problem by first drawing a diagram and identifying any variables that we're going to be working with.

Since this is a rectangular fence, I'm going to go ahead and draw a rectangle to represent that. I know that one side of this rectangle is formed by a rock wall, so I only need to focus on these three sides here. In terms of variables, I can express the length and the width of this rectangle as variables. It doesn't matter what those variables are. Here, I'm going to use x and y.

Now with step 1 done, we can proceed to step 2, where we want to write a function for the value being optimized. We already recognized here that we want to optimize the area. That means that we want to write a function for the area. We know that the area of a rectangle is just the length times the width. Based on the variables we have here, that would be x times y.

We're not quite done yet because we have this function but want it in terms of just one variable. This will often require additional equations. Where do we get those equations from? Well, this is where our constraints come in. Remember, in our problem, we were told that we only have 200 feet of fencing.

These three sides can only be made up of 200 feet. In other words, the perimeter of this fence is 200. Based on the variables that we have here, if I add all those side lengths up to get my perimeter, that means that 200 is equal to x + x + y, those 3 side lengths added up together. If I go ahead and combine like terms here, I can simplify this a little to 2x + y. How can we use this to put our area function in terms of just one variable?

Well, if I subtract 2x on both sides here, this ends up giving me y = 200 - 2x. I can then substitute y in my area function here to get it in terms of just x. So that would mean that a=x×200−2x. This is now just in terms of x. I can distribute this x into my parentheses to give me an area function of 200x - 2x².

So I have my area function now just in terms of x. Step 2 is completed, and we can move on to step 3. Now in step 3, we want to determine the domain restrictions on our function. Looking at this function, you may be thinking, this is a basic polynomial. So isn't its domain just all real numbers?

You would be right if this polynomial was just out in the wild and didn't have any scenario attached to it. But here, since we're working specifically with the dimensions of fencing and area, there are constraints on this value. Thinking about what's going on here, I know that I can't work with a negative dimension of fencing. My length can't be negative. So I know that this value x must be greater than or equal to 0.

If I look further at my diagram here, knowing that I only have 200 feet of fencing to work with, if I use up all that fence on my two side lengths x, that means I put 100 feet on this side, 100 on the other side, leaving nothing for y. I've used up all 200 feet of my fencing. If I made x any greater than 100, I wouldn't have enough fencing to do so. So that means that x must also be less than or equal to 100. So I have my domain restriction here that x must be greater than or equal to 0 and less than or equal to 100 based on the specific scenario that we're working with here.

Now we can move on to step 4, and that's where this process is going to start to feel a bit more familiar. Here we're going to find our critical points, and when working with these problems, that means that we just want to find where that first derivative is equal to 0, as we've done a million times before. So we want to first find the derivative of our area function. That's a prime. Using the power rule here, that's going to be 200 - 4x.

Now we want to set this equal to 0 to find any critical points. If I move 4x over to the other side, I get 200 equals 4x. And then if I divide both sides by 4, that gives me a critical point at x = 50. Now from here, we can move on to step 5. And when getting into step 5, we actually want to think back to our domain restriction.

Here, we know that x should be greater than or equal to 0 and less than or equal to 100. These 0 and 100 represent endpoints. Since here our function has endpoints, or in other words, we're working with a closed interval, that means

Example 1: Minimizing Surface Area

Here, I'm going to walk you through solving a really common type of applied optimization problem. We're tasked with minimizing the surface area of a three-dimensional object. Now we're going to follow the same process that we just learned in order to solve this problem. So let's go ahead and get started. Here, we're told the shipping company wants to construct a rectangular box with a square base to have a volume of 125 cubic inches.

If the box has no lid, what should the dimensions of this box be to use the least amount of cardboard? Now we're not explicitly told here, hey. Minimize the surface area of this box. What we're asked to do is to find the dimensions to use the least amount of cardboard. Now, the least amount of cardboard translates to having the least amount of surface area for this box.

That's why this is really a problem of minimizing the surface area. So now let's go ahead and get started with our first step here and draw a diagram and identify any variables we'll be working with. Now we already have an image to represent our problem here. We have a rectangular box with a square base, so let's think about any variables we should be using here. Now here we're told that this is a square base, meaning that both of these sides are the exact same.

So I'm going to refer to both of these sides as x. Then I'm going to refer to the height of this box as y. Okay. With those variables identified, let's move on to step 2 and write a function for our value being optimized. Now here, we want to minimize the surface area.

So that means that we want to write a function to represent that surface area so that we can then minimize it. So, writing a function for our surface area here, we know that the surface area of a box is just all of those sides added up together. Now four of our sides are going to have an area of x times y. So I have 4xy plus the area of my base here, which is x times x or x2. Now remember that this box has no lid, so that means that I don't have to account for another base there with my surface area.

This is my function representing my surface area. Remember, we want this to be in terms of one variable, which often requires an additional equation. Remember that these equations are typically based on the constraints given to us in our problem. In our problem, we're told that this box has a volume of 125 cubic inches. So v is equal to 125.

Now looking at my box, I can determine my volume by multiplying all of those sides together, length times width times height. So that's going to end up being x times x times y. Now x times x is just x2, so we can simplify this a little bit. So 125=x2×y. Now if I were to divide both sides by x squared, the x squared will cancel, and this then tells me that y is equal to 125x2.

Now that I have y in terms of x, I can substitute this back into my surface area function in order to get my surface area function in terms of just one variable, which is exactly what I want. So substituting that y and bringing my surface area equation down here, this is 500x+x2. So now I do have my function in terms of just one variable, so we can move on to step 3 and determine our domain restrictions. Now in this problem, the only information that we are given is that our volume is 125 cubic inches. We're not given a specific amount of cardboard that we're working with because that's what we're trying to determine, so I don't really know a ton of information that's restricting this x value. What I do know is since I'm working with the dimensions of a box, this value cannot be negative.

So I know that x must be greater than or equal to zero. Now I don't know anything else about that, so this is my entire domain restriction. So we can move on to step 4 now. Remember, this is where our steps may start to feel a bit more familiar to what we've worked on in the past. Now here, we want to determine our critical points, specifically where that derivative is equal to zero.

So let's go ahead and find the derivative of our surface area function. So that's a prime. Now finding our derivative here, this will be -500x2+2x. Now we want to set this equal to zero here and solve for x. So solving for x here, I'mgonna go ahead and pull one of these to the other side.

So that's 500/x2=2x. If I multiply both sides by x squared to get rid of that on the bottom there, that will cancel, and I'm left with 500=2x3. Now if I divide both sides by 2 here, that gives me that 250=x3. Then finally, cube rooting each side.

My x is going to be equal to the cube root of 250, simplified in its simplest terms using that cube root. This is going to be 5×2, or as a decimal, this works out to be about 6.3. So this is about 6.3 just because we're doing a little bit of rounding there. Okay. So we have our only critical point for step 4.

Now we can move on to step 5. This is where we're going to split off based on whether we have endpoints or not. Now we know that our domain restriction told us that x could just be greater than or equal to zero. That means we don't have another endpoint that we're working with, so we're working with an open interval. There are no endpoints here.

So because of that, we want to proceed with using our second derivative test. In order to do that, we're just going to plug our critical point into that second derivative and pay attention to the sign. So the first thing that we need to do here is to figure out what that second derivative is. Now remember that our first derivative was right here. It's -500x2+2x.

To get our second derivative, we're just differentiating that again. So let's go ahead and find that second derivative here, a double prime. Differentiating it again here, this will work out to be 1000x3+2. Now looking at this, since using our second derivative test, we're paying attention to the sign of our second derivative. Looking at this value, knowing that all of our x values must be greater than or equal to zero, I know that no matter what x value I plug in here, this second derivative will always be positive.

So that means that because this second derivative is positive everywhere, so if I were to plug that critical point in, I would get a positive value, that tells me that this critical point is indeed a minimum. So this critical point that I found gives me my minimum surface area. So what is that minimum surface area? Well, in order to get that, we need to plug back into our original function. Remember to stay organized when working through these problems because I know there's a lot happening here.

So we have determined that this critical point results in our minimum. So we do have one dimension of our box already, this 6.3 inches that represents the sides of the base. So this is 6.3 inches. Now in this problem specifically, we're just asked to find the dimensions. So I don't actually even have to find what that minimum surface area is unless I really want to.

What I really want to find is what this y value is so that I can have all of the dimensions of my box. So from here, I want to go ahead and go back to my y equation, 125x2, and I want to plug that 6.3 inches into that equation for y so that I can get that final dimension there. So this is going to be y equals 1256.32.

Now when I work this out, this will give me about 3.15 inches for that final dimension of my box here. I can go ahead and label that 3.15 inches. So, based on everything that we found here, the dimensions that will give me the minimum surface area that will use the least amount of cardboard here are a height of 3.15 inches and a base of 6.3 inches on either side since that is a square base. So we've successfully solved this problem. I know that these problems can feel a little bit intimidating and sometimes a bit long.

I know that there's a ton of algebra and math going on here, but you're doing a great job. Now we're going to continue practicing with more applied optimization problems coming up next. I'll see you there.

A poster is set to have a total area of 1150 cm², with 2-cm margins on the sides and the top, and a 3-cm margin at the bottom. What dimensions will maximize the printed area?

• A. Width: 22.5 cm, Height: 51.2 cm
• B. Width: 35.6 cm, Height: 32.3 cm
• C. Width = 30.3 cm, Height = 37.9 cm
• D. Width: 28.1 cm, Height: 41.0 cm

Maximizing Profit & Revenue

Throughout the past couple of videos, we've been learning how to optimize different real-world values, and we can do this by expressing that value as a function and then finding the maximum or minimum. Now here, we're going to continue working through applied optimization problems, focused specifically on maximizing the revenue or profit of a business. Now, we can solve these problems the exact same way that we have any of our applied optimization problems, just with a couple of new equations in mind. Now here, I'm going to walk you through how to set up these problems dealing with revenue or profit. So let's get right into our example here.

Here, we're told that a coffee shop sells x coffees a day at p dollars per coffee. Now we're given a price demand function here. P of x, our price, is equal to px=100-x4, and we're asked to find how many coffees the shop needs to sell in order to maximize their daily revenue. So since we're asked to maximize the daily revenue, that means that we want to express revenue as a function and then find its maximum. Now we know exactly how to do this, so let's get into our steps here.

Our first step is to draw a diagram and identify any variables. Now when working with these problems with revenue or profit, a diagram is not usually going to be helpful for us because we're not dealing with the dimensions of anything. And in terms of variables, our variables are typically going to be given to us in our problem. Like, here, we're told that our coffee shop sells x coffees a day at p dollars per coffee. So we can go ahead and jump to step 2.

This is where we want to write a function for our value being optimized. Now we know here that that is revenue. So we want to write a function to represent revenue. Revenue should always be equal to the price of your items multiplied by the number of items sold by your business. Now this makes sense because the amount of money you make, your revenue, should be equal to the price of each of your items multiplied by the number of items that you actually sell.

So for this business specifically, our revenue function for this coffee shop is going to be equal to that price, which will typically be a function of x given to you in your problem here as it is here. So this is 100 minus x over 4 plugging in that price function multiplied by x. Now if your item sold is ever something different than x, this will typically be expressed to you in your problem. But most times, it will just be x. So we have our revenue function here.

We can go ahead and distribute that x into our parentheses. That gives us Rx=100x-x24 for the revenue of this coffee shop. Now we can see that this is already just in terms of x, so we don't have to worry about doing anything else to get this in terms of just one variable. Now if instead here we were asked to maximize the profit of this coffee shop, we would want to take that revenue function that we just found, the amount of money that we got from selling all of those coffees, and subtract the cost of actually operating our business. This would give us our profit.

But here, we're just working with revenue, so we can go ahead and proceed on to step 3 and determine any domain restrictions. Now in terms of domain restrictions here, since x represents the number of coffees that I sell a day, I know that I can't sell a negative number of coffees. So I know that x must be greater than or equal to 0. Now in terms of anything else going on in my problem, I know that I also can't have a negative price. So I know that that price function, p of x, must also be greater than or equal to 0.

And since I know exactly what that function is in terms of x, I can go ahead and plug that in. 100 minus x over 4 must be greater than or equal to 0 because I can't put a negative price tag on those coffees. Now from here, if I solve for x in this inequality, this results in x being less than or equal to 400. So with these two domain restrictions in mind, I know that x should be greater than or equal to 0 and less than or equal to 400. So step 3 is completed.

From here, we would proceed in solving this the same way we would any applied optimization problem, finding our critical points and then using either our extreme value theorem or our second derivative test to determine our maximum. Now you should go ahead and try to finish solving this problem on your own. In the next video, you can finish solving it alongside me. I'll see you there.

Example 2: Maximizing Revenue

If you haven't already, go back to the previous video to see exactly how we set this problem up. At this point, we already have our revenue function, and we have our domain restrictions. And we're moving on to step 4, finding our critical points. Now, just as a recap, remember that our revenue function is given by the price times the number of items sold. And here, that ended up being R(x)=100x-x24 with our domain, such that x≥0 and x≤400 based on the fact that we have to sell a number of items that is non-negative and our price must also not be negative.

So from here, jumping into step 4, we're going to find our critical points. That's where our first derivative is equal to 0. So we need to find the derivative of our revenue function, which is going to be R'(x)=100-2x4, which simplifies to 100-x2. Now we want to set this equal to 0 to find our critical points.

If I move that x2 to the other side, I get 100=x2. Multiplying both sides by 2 gives me then that x=200. Now based on this domain restriction here that x≥0 and x≤400, as we move into step 5, we do have endpoints here working with a closed interval, meaning we can find our maximum value by plugging all of these values in, our 2 endpoints and our critical point, to our original function. So we want to get R(0), R(200), and R(400). That's our 2 endpoints and our critical point that we just found.

Now plugging these into our original function, that's R(x)=100x-x24. If we plug 0 in, this will actually work out to be 0. And if we plug our other endpoint 400 in, that will also end up being 0. Now if we plug 200 in, this will be 100×200-20024, which gives us a value of 10,000. Now we've seen this happen before that our endpoints work out to be 0, so it then becomes obvious that our largest value here is created by our critical point here.

So this represents our maximum. Now based on what we're looking at here, this is our maximum revenue, so our maximum revenue is $10,000. In our original problem, we were asked how many coffees do we need to sell in order to get that maximum revenue. Now we plugged in a critical value of 200 in order to get this maximum, and that x represents the number of items that are sold. So we need to sell 200 coffees in order to reach our maximum revenue of $10,000.

Now something that you may see either in your textbook or done by your professor in your classroom is instead of going ahead and plugging in your endpoints along with your critical point to that original function, even if you have endpoints, some people may choose to default to the second derivative test and just go ahead and find that second derivative and determine whether it's positive or negative. If we were to have done that here, our second derivative, that's R''(x), would have ended up being negative one half. Now we can see that this value is clearly always negative, meaning that we know further that this point is a maximum. Now this is definitely an acceptable method to do as well. Feel free to ask us if you have any questions, and I'll see you in the next one.

Example 3: Maximizing Profit

We just learned how to write the equations for revenue and profit. Because of that, we can use our same process that we use for any applied optimization problem when dealing with revenue or profit. Now, here, we're going to put that knowledge to the test faced with a problem of where we want to maximize the profit of a business. So let's go ahead and get started. Now in this problem, we're told that Gerald and Katya recently started selling custom-made phone cases, and they've noticed that the price per case they can charge depends on how many cases they sell, which is represented by this function here, p(x) = 50 - 0.2x.

So this is our price per case. X is the number of cases sold. And as I just said, p(x) is the price per case specifically in dollars. Now on top of that price per case, the cost for making x number of cases is given by this cost function here, 100 + 15x, where that is the total cost in dollars to produce x cases. Now, of course, these two want to maximize their profit.

So here, we want to find a couple of different things. We want to find the number of cases that should be sold to result in that maximum profit, what the maximum profit is, and the corresponding price that these two should sell their phone cases at. Now because of this, we want to maximize the profit. Right? So we need to write a function to represent that profit.

Now looking to our steps here, we know that we already have our variables identified and there is no need for a diagram here, so we want to start by just writing the function for our value being optimized, which we know is the profit. Now remember that the profit of a business, so that's capital P(x), this is going to be equal to the revenue minus the cost. Now we're given that cost function explicitly here, 100 + 15x, but we actually have to come up with what the revenue function is before we can make our profit function. Remember that the revenue is just equal to the price times the number of items sold, which in this case is x. So our revenue is just going to be equal to that function p(x) multiplied by x.

So we can go ahead and set up our profit equation here. So P(x), our profit, is going to be equal to (50 - 0.2x) × x, this represents our revenue, minus the cost of making these phone cases, 100 + 15x. Now we can do some simplification here by distributing this x and also combining like terms by distributing that negative. This results in our profit equation being P ( x ) = - 0.2 x 2 + 35 x - 100 .

Now remember that, as I said before when we were talking about revenue, we don't want to sell a negative number of items. Right? So we know that x must be greater than or equal to 0. Now on top of that, we also can't sell things at a negative price. We're not going to pay people money to come get our phone cases.

Right? So we know that p(x), this function up here, must also be greater than or equal to 0. Now when we solve this inequality, this results in telling us that x should be less than or equal to 250. So bringing that domain restriction down here, combining it with the fact that x should be greater than or equal to 0, it should also be less than or equal to 250. So with those domain restrictions, let's go ahead and proceed in finding our critical points.

That's where that first derivative is equal to 0. So when we do that, we want to find that first derivative P'(x). We can do using the power rule. So this is going to be a -0.4x + 35. And we, of course, want to now set this equal to 0.

When we do that, we can move that -0.4x to the other side. This gives us 35 = 0.4x. Solving for x here by dividing both sides by that 0.4, this results in an x value of 87.5. Now something to keep in mind here is that we cannot sell half of a phone case. Right?

So because of that, we want to round this to the nearest whole number. So instead of 87.5, we want to work with here 88 because it doesn't make sense for us to sell half a phone case. I'm not going to break it in half and sell that to a customer. So with these 88 phone cases in mind, let's move on to step 5. Now as I mentioned in the previous video, some people choose to default to the second derivative test, especially when dealing with such a simple first derivative, meaning that it would be really simple to find the second derivative.

Now either method that you choose here will result in the same exact answer. So I'm going to go ahead and use my second derivative test here even though I have endpoints. You can choose to use those endpoints in order to use your extreme value theorem. Now finding my second derivative here, that's p double prime of x, that's going to be equal to -0.4 based on what our first derivative test is right So this is what I meant about our second derivative being easy to find here. Now because this doesn't have any variables, this tells us that our second derivative is always negative.

When our second derivative is negative, according to our second derivative test, that means that our function is concave down, and that point that we're working with is a maximum. So we know that these 88 phone cases being sold result in a maximum. Let's make sure we're answering all of the questions being asked of us. Now we want to know the number of cases that should be sold to result in that maximum profit. We found that right here.

It is 88 phone cases. Now we also want to determine what exactly that maximum profit is that we get from selling those 88 phone cases. So in order to find that, we can plug 88 back into our profit function. So we can find P(88). This is equal to -0.2 times 88 squared + 35 times 88 - 100.

Now when we plug this in, this will result in $1,431.20 because this is our profit in dollars. So this is our answer to part b, $1,431.20 having sold those 88 phone cases. Now what price corresponds to this maximum profit? How can we get that? Well, in our original problem, remember that we were told the price is 50 - 0.2x.

So all we have to do here is plug that 88 into our price function as well. So this is a lowercase p as opposed to capital P for profit. So this is lowercase p(88), and that is equal to 50 minus 0.2 times 88, which results here in a price of $32.40. So our answer for part c here, that corresponding price that gives us our maximum profit, that is $32.40. So we have successfully solved this problem using everything that we know about applied optimization.

Feel free to let us know if you have any questions here, and I'll see you in the next one.

Your café sells lattes for $4 each to 100 customers per day. For every $1 increase in price, you would lose 20 customers. Find the price that maximizes revenue. Hint: The # of items sold is based on the number of customers.

• A. $4.00
• B. $4.50
• C. $5.00
• D. $5.50

Minimizing Inventory Costs (Inventory Control)

Throughout the past several videos, we've been working to optimize different real-world values, and we're going to continue along that path here. Here, we're going to specifically focus on minimizing the inventory costs of a business. This specific type of applied optimization problem is often referred to as an inventory control problem. We are going to solve these problems the same exact way that we have all of our applied optimization problems, just with a new equation in mind. Now I'm going to walk you through how exactly to set these problems up in this video.

So let's get into our example here. We're told that a math store sells 400 calculators per year. It costs $2 to store one calculator for a year, and to reorder, there is a fixed cost of $25. Now we want to find how many times a year this company should order calculators in order to minimize their costs. We want to minimize the total costs associated with this business.

Based on this problem, we can see that there are costs associated with storing our items and with reordering them. So here, we want to minimize the total cost of both storing and ordering inventory. In our very first step, we want to draw a diagram and identify any variables that we're going to be working with. As it was with profit and revenue, it's not going to be super helpful to draw a diagram here, but I do want to think about what variables I'm going to be using.

In this problem, we're told specifically to let \( x \) represent the number of items per order. Moving on to step 2, we want to write a function for our value being optimized. We know that that's the total cost. Our total cost is going to account for both the cost to store our items, so our storage cost, and the cost associated with reordering them.

We need to think about both of these things individually and add them together to get that total cost. Looking at the storage cost first, the total cost of storing your items is going to be equal to the storage cost per individual item multiplied by the average number of items that you have in storage. The average number of items in storage throughout the year averages out to be \( x/2 \). So if I take those $2 and multiply it by the average number of items in storage, this gives me my storage costs.

Now we want to add this together with our reorder cost. Your reorder cost is going to be your cost per individual order multiplied by the number of times you order throughout the year. The number of orders is going to be the total items that you expect to sell divided by the number of items per order, and we know this is \( x \). So, the cost per individual order, a fixed cost of $25, can be shown in the cost function as: C ( x ) = x + 10000 x

This is the most common way that you should go about setting these problems up. This will work most of the time, but remember that some problems may present information to you slightly differently. You should always pay attention to the exact problem that you're working with, and make adjustments accordingly. With this cost function in mind, let's go ahead and move on to step 3 and determine our domain restrictions.

Here, \( x \) represents the number of items per order. We can't order a negative number of items, nor can we place an order for 0 items. So \( x \) must be at least 1, so it should be greater than or equal to 1. Because we only expect to sell 400 calculators per year, \( x \) must also be less than or equal to 400. From here, you can proceed in solving this problem as you would any applied optimization problem by finding your critical points and applying either your extreme value theorem or your second derivative test. As you continue solving this problem, remember that \( x \) is the number of items per order, and you cannot order half of a calculator or half of any item. If you end up getting a decimal value for \( x \), you should test your nearest whole numbers in order to get your final answer.

Now, you should go ahead and try to finish solving this problem on your own, and you can check in with me in the next video to check your answer. Thanks for watching, and I'll see you there.

Example 4: Inventory Control

If you haven't already, check back to the previous video to see exactly how I set up this problem. At this point, I already have my cost function that I want to minimize and my domain restrictions so I can proceed to step 4 and find my critical points. Now, to do this, I need to take the derivative and set it equal to 0. First, I need to find what my derivative is. So \( c'(x) \) here, this is going to be \( 1 - \frac{10,000}{x^2} \).

Now, setting this equal to 0 here if I move that \( \frac{10,000}{x^2} \) to the other side, this becomes \( 1 = \frac{10,000}{x^2} \). Then if I multiply both sides by \( x^2 \), this gives me that \( x^2 \) is equal to 10,000 because the \( x^2 \) will cancel and \( x^2 \times 1 \) is just \( x^2 \). From here, I can take the square root of both sides, which gives me that \( x \) is equal to plus or minus 100. Now thinking back to my domain restriction here, I know that \( x \) has to be between 100 and 1400, which throws negative 100 out of the mix. So I only need to worry about my critical point \( x = 100 \).

With this critical point found, I can proceed to step 5. Because I have endpoints here, I'm working with a closed interval. I can go ahead and use the extreme value theorem, plugging in both of my endpoints 1400 and my critical point 100 to my original function. Because we're looking to minimize the cost, we're going to be looking for our smallest value when we do this. So let's go ahead and plug these back into our original cost function.

So both of my endpoints here, I want to find \( c(1) \) and \( c(400) \) and my critical point \( c(100) \). Just putting those in numerical order. Now, remember, we're plugging into our original cost function, not our derivative. So this is \( 1 + \frac{10,000}{1} \). This is of course 10,001.

And then if I plug 100 in, this is \( 100 + \frac{10,000}{100} \). This works out to be 200. Then plugging in my other endpoint, this is \( 400 + \frac{10,000}{400} \), which is 25. So this is 425. Again, we're looking for our smallest value here.

And of these three values, I can see that 200 is my smallest value. Now let's think about what's actually going on here. It's important to pause sometimes and really pay attention to what's happening. This is saying that if we only ordered 1 item per order, that means we're gonna have to order 400 times a year to keep up with our demand of 400 calculators and it would cost $10,001 to do that.

It makes sense that the highest cost is associated with ordering 1 item at a time. Now this is if we ordered all 400 calculators at once. And while this is a lower cost than ordering 1 at a time, it is still not quite as low as splitting up our orders into 100 items per order. Now in our original question, we were specifically asked how many times a year our company should order calculators to minimize their cost. What I just found here was the cost associated with ordering 100 calculators per order.

So we need to go ahead and translate this to thinking about what our actual number of orders is. So if I'm ordering 100 calculators per order and I want to order a total of 400 calculators throughout the year, I can take 400, that total, and divide it by the number of items per order. That's 100. This is equal to 4. Remember, this number of orders is exactly what we plugged in when calculating our reorder cost. So this problem, we should order 4 times a year to stack our math store most efficiently.

Now if you have any questions here, let me know, and we're going to get a bit more practice with inventory control coming up next.

A small clinic uses 5,000 bottles of hand sanitizer per year. Each shipment costs $200 to order, and it costs $1 to store each bottle for a year. How many bottles should the clinic order in each shipment to minimize the total ordering and storage costs?

• A. 1414 bottles
• B. 2512 bottles
• C. 1008 bottles
• D. 943 bottles

Example 5: Norman Window

In the past couple of videos, we've been looking at some common business applications of applied optimization problems. Here, we're going to continue working with some more common applied optimization problems, starting here with the Norman window. Now the Norman window is a common problem type that you'll encounter just based on the specific shape of the window. A Norman window is shaped as a rectangle with a semicircle on top. Now in this problem, we're going to be following the exact same process that we have been to solve all of our other applied optimization problems.

So let's go ahead and jump in here. Now like I said, our Norman window is shaped as a rectangle with a semicircle on top. We're told here that the perimeter of this window is 24 meters and this semicircular part lets in 1 quarter as much light per square meter as the rectangular part. We're asked to find here the dimensions of the window, aka the width and the height of the rectangle, that allow the maximum amount of light to pass through, considering that window trim negligible. So we don't have to worry about the width of that.

Now here, we're asked to find what will allow the maximum amount of light. And this might seem kind of strange because we don't have an equation in calculus for light that we can then maximize. So how exactly are we going to go about this? Well, because here, we can think of this as the area of a window, if we maximize the area, it will let more light in. We can effectively write an equation that represents the area and that will maximize that light.

So here, even though we're asked specifically to maximize the light, really, we're just maximizing the area. Now we're also told in this problem that the semicircular part allows a 1 quarter as much light in as the rectangular part, and this is something that we'll need to consider when we're writing our equations for that area. So let's go ahead and start with step 1 here and draw a diagram and identify any variables. Now here, I already have my Norman window drawn for me, and I know that this side length is going I'm gonna call that x and this side length I'm going to refer to as y. Now because we're working with this semicircle up top, when we're working with circles, it is important to pay attention to the radius there.

And because I know that this side length is x, that makes this part x as well. So that radius will then be x over 2. So with those variables identified, let's move on to step 2 and start writing our function for our value being optimized. Now as I mentioned earlier, even though we're asked to maximize the light effectively, we're just going to be maximizing the area with some special considerations taken. So here, we wanna determine what the total area of our window is.

Now, in order to do that, we need to account for both the rectangular part and the semicircular part. So that total area, A, is going to be the sum of the area of that rectangle plus the area of that semicircle. Now, even though we're working with the area here, since our ultimate goal is to determine what will allow the maximum amount of light, we need to take into account the fact that our semicircle is only letting in one quarter of the light of that rectangle. So because of that, when we write this equation, we wanna add the area of the rectangle plus 1 quarter of the area of that semicircle. So let's go ahead and write our equations here.

We know that the area of a rectangle is just the length times the width. And here, that's just x times y. So we have x times y plus that 1 fourth times the area of that semicircle. The area of a circle is πr2. This is only half of a circle, so that's going to be 1 half πr2.

We know that our radius here is x over 2. So this is 1 half times πx22. Now we want to do some simplification here. This is just x times y. That's rather simple.

But we have these fractions, 1 fourth and one half, and we also have this x over 2 being squared. That gives us x squared over 4. So on the top of my fraction, I have put πx2. On the bottom, I have 4 times 2 times 2 squared which gives me 32. So this is my area function with the fact that our semicircular part lets in a 1 quarter of the light accounted for by that 1 fourth.

So we have our area equation here, but remember that we want it in terms of just one variable, so we need to figure out how to get rid of that y. Remember that π is not a variable. It's a constant, so that is totally fine to be in our equation here. So in order to get an equation for y that we can then replace in here, we wanna look to what constraints are given to us in our problem. And here, we're told that the perimeter of our window is equal to 24 meters.

So the perimeter I know is all of my sides added up together. I know that that's x plus y plus this other side of y as well. And then I also need to account for that semicircle. So 24 is equal to x plus y plus y, which I can write as 2y, plus the perimeter of this circle, which you may also know as the circumference of a circle. Remember that it's only half of a circle, so even though the circumference is 2 πr, half of this would just be πr.

So I want to take π and multiply it by the radius, which is x over 2. So I can rearrange this equation now just for y. Remember, since I'm working with the perimeter here, in this case, I do not need to account for the fact that that semicircle is only letting in one quarter of light because that has nothing to do with how much window trim is around there. So if I go ahead and subtract x and π x over 2 from both sides here, that gives me that 2 y, not forgetting that 2, is equal to 24 minus x minus π x over 2. Now dividing everything there by 2 to get y all by itself, this gives me that y is equal to 12 minus x over 2 minus π x over 4.

Now we can do some simplification here, but I'm gonna save that for actually plugging it into my area function here. So let's go ahead and plug that in. So a is equal to x times y, which is what all of this is here. 12 minus x over 2 and minus π x over 4. Then not forgetting everything else in my area equation, I have this second term, πx2 over 32.

So this is now in terms of one variable, but we know that there's some simplification to be done. Let's go ahead and distribute this x into my parentheses. This will give me 12x minus x squared over 2 minus π x squared over 4 plus πx2 over 32. Now we see that all of these terms have an x squared in them, so that means that we can go ahead and factor out that x squared. This will give us our area function as 12 x.

I'm gonna factor out actually positive x squared and keep those constants negative. So this is negative one half minus π over 4 plus π over 32. Now I can go ahead and combine these terms, π over 4 and π over 32, just to make this a little bit cleaner. So this is 12 x. I'm just going to go ahead and combine it in my parentheses, negative π over 4 and a positive π over 32 will give me negative 7 π over 32.

So this is simplified as much as I can without getting rid of anything else because remember we don't want to go ahead and multiply everything out here because that means we'll have to round. We don't want to round until it's necessary. So this is our area function here 12 x plus x squared times negative 1 half minus 7 π over 32. From this point, we wanna determine our domain restrictions. Now I know that this can often be a sort of tricky part of these problems, so let's go through this together.

Now I know that in terms of dimensions, I know that my x value cannot be negative. So I know that x must be greater than or equal to 0. But what does it need to be less than or equal to from, if anything? Well, here, we're limited by the perimeter of our window. I know that the perimeter of my window can only be 24 meters.

So if I were to use all of that material on my perimeters involving x, so that's gonna be this x on the bottom and then the fact that I have the top part of my window here. I know that the perimeter up here is π x over 2. So if I'm using all of my material, 4x+2, and all of my material is 24 meters, this tells me that x would then be equal to about 9.34. So the fact that using all of my material means that x would be equal to that 9.34, this gives me my other endpoint here that x has to be less than or equal to 9.34. If it's any greater than that, then I won't have enough material.

So now that I have my domain restrictions here, x must be greater than or equal to 0 and less than or equal to 9.34. I can go ahead and proceed into step 4 and find my critical points where that derivative is equal to 0. So let's go ahead and find the derivative of our function here. That is going to be a prime. Looking at our area function, our derivative, even though this looks rather complicated, we're still just using the power rule.

This is going to be 12 plus 2x times all of this because it's just constants. So negative 1 half minus 7 π over 32. Now we wanna go ahead and set this equal to 0 to get those critical points. If I move that 12 over to the other side, it will cancel and I need to divide definitely by that 2. This will give me x times all of this negative one half minus 7 π over 32 is equal to negative 6.

Now the last step here in isolating x is where we're going to want to go ahead and round because we are going to want to plug this into our calculator to divide a negative 6 by it. So we're going to divide by negative 1 half minus 7 π over 32 in order to get that to cancel on this side and our answer will be negative 6 over negative 1 half minus 7π over 32. You're going to want to go ahead and plug this into your calculator. When you do plug this into your calculator, you'll get that x is about 5.05. So I have my critical point here.

And looking back at my final step because that's where I'm at now, step 5, because I do have endpoints, I know that x should be greater than or equal to 0 and less than or equal to 9.34. I am working with a closed interval. So I'm going to go ahead and use the extreme value theorem, plugging my critical points and my endpoints back into my original function. And since I'm looking to maximize this value, I'm going to be looking for the largest value here. So let's go ahead and plug these back into our original function.

I'm going to go ahead and do that over here on this side. So plugging my endpoints in here, that's a of 0 and a of 9.34 and then my critical point, a of 5.05, just putting those in numerical order. Now you're gonna wanna go ahead and use a calculator for this because we are working with pies in our original function. Remember, we're plugging this into just a, not a prime, our derivative. Make sure to keep this organized.

So we're plugging into 12x + x squared minus one two - 7 π over 32. You are going to want to use a calculator for this. When you plug 0 in to that function, you will end up with 0. When you plug 5.05 in, you'll end up with about 30.32. And when you plug 9.34 in, you'll end up with 8.51.

Now remember, we're trying to maximize this value, so we want to look for the largest one. That is this 30.32 from plugging that 5.05 in. That was my critical point. Because plugging that critical point into my original function resulted in the largest value, That means that one of my dimensions that creates this maximized light passing through is this x equals 5.05. And this is in meters because those are the units being used in our problem.

So I can go ahead and label that dimension here. I know that this x is equal to 5.05 meters. And that means that this radius here will just be that 5.05 divided by 2. But it remains here to figure out what y is. So remember that earlier we did have an equation for y.

We know that y is equal to 12 minus x over 2 minus π x over 4. So we can plug that 5.05 in to get what this dimension should be. So y equals 12 minus 5.05 I'm plugging this x value in over 2 minus π times 5.05 over 4. This results in a y value rounded to about 5.51 meters. So this gives me my other dimension here that this side length should be 5.51 meters.

So in order to maximize the light passing through my window, based on the fact that the rectangular part is going to let more light pass through, in fact, 4 times as much light through because this semicircular part only lets in one quarter of the light, my dimensions should be that this side length is 5.05 meters and this side length is 5.51 meters. Now we're going to continue getting practice with some more common applied optimization problems coming up in the next video. I'll see you there.

Example 6: Packaging Design

Here, we're going to look at another type of common applied optimization problem where we're tasked with designing some sort of packaging. Here, we're going to be working with a box, and our goal is to maximize the volume of this box. When working with these packaging design problems, you may be maximizing the volume or minimizing the volume or maximizing or minimizing something else entirely like the surface area. Remember, for all of our applied optimization problems, we still want to work through our steps in the same exact way. So let's get into this problem.

Here, we have a box with an open top that is formed by cutting equal-sized squares from the corners of a 3-foot by 8-foot rectangular sheet of cardboard and then folding up the sides. We want to find here the dimensions of our cut squares that will give the box our maximum possible volume. We also want to state what that volume is. Ultimately, because we're trying to maximize the volume, that means that we want to write a function that represents that volume and then use the strategies that we know to maximize it. So, for step 1, we want to draw a diagram and identify our variables, and we can see that that is already done here.

We have our rectangular sheet 8 foot by 3 foot. And because these are squares cut out from the corner, I know that all of those sides are equal, so I've labeled them both as x. So with step 1 done, we can proceed in actually writing a function for our value being optimized, which here is the volume. Now we have to be careful here, even though we know that the volume is length times width times height. Looking at this 8 and this 3, we can't just take 8 times 3 and multiply it by our height of x because we're cutting these corners out.

So when writing our volume function here, we need to account for that. So this first side length of 8, I need to subtract these 2 little squares being cut out. Each of them has a length of x. So if I subtract 8 minus x minus x, that is 8 minus 2x. Now I need to do the same thing with my other side length here because I need to account for these little squares.

This is going to be 3 minus 2x. Then we need to multiply this, of course, by the height as well, which will end up being x when those sides get folded up. So from here, we can go ahead and foil this out and distribute that x, and this will end up giving us 24x−16x2−6x2+4x3 using our foiling method. Now I can go ahead and combine these like terms here and then rearrange this slightly to give me 4x3−22x2+24x, just writing this in standard form. So here is my volume function.

And conveniently, it's already in terms of one variable because we only have one variable, x. Now if these were not squares and they were actually rectangles being cut out of our sides, we would have another variable to account for. But because here we don't, we can go ahead and move on to step 3 and determine our domain restrictions. Now because here we're working with dimensions, we know that x must be greater than or equal to 0.

But is there another restriction on this domain? Well, we know that these side lengths are 3 by 8. Looking at my smallest side of 3 feet, I could cut that 3 feet in half to use to cut the maximum amount out of these corners here. If I entirely just cut out a square of 3 by 3 here, that's the furthest I could go because this side is only 3 feet long. So if I take 3 and divide it by 2, that gives me the maximum amount I could cut in here.

3 divided by 2 is 1.5, so x must be less than or equal to 1.5. So with that domain restriction in mind, we move on to step 4 to find our critical points. This is where our first derivative is equal to 0. So v prime here. Using the power rule, that's going to give me 12x3−44x+24.

Now I can do a little bit of factoring here. If I go ahead and factor a 4 out of all of these terms, that's going to give me 4 times 3x2−11x+6. Now we want to set this equal to 0 because that 4 will just cancel when I do that. I don't have to worry about it. From here, you can use whatever method you prefer for solving quadratic equations.

Because here, I have this first term of 3, I'm going to go ahead and use the quadratic formula so I don't have to worry about any of my other methods here. So using my quadratic formula remember our quadratic formula tells us that x is equal to negative b plus or minus the square root of b squared minus 4ac all over 2a. Now if you don't know the song for this to remember it, I encourage you to look that up. But from here, let's actually plug in what these values are. Remember, this a, b, and c just corresponds to our coefficients here.

So we have a, b, and c. So plugging those in, negative negative 11 will give me positive 11. So 11 plus or minus the square root of that negative 11 squared minus 4 times 3 times 6, all over 2 times 3. Now this simplifies quite a bit. This ends up being 11 plus or minus the square root of 49 divided by 6.

We know that the square root of 49 is just 7. So this is 11 plus or minus 7 over 6. Now if we work this out to our two possibilities here, so if we take 11+7 over 6 and 11 minus 7 over 6, This gives us 2 possible critical points at 3 and at 2 thirds. Now looking back to our domain restriction here, we know that this value can only be between 0 and 1.5, so we don't have to worry about that first critical point of 3 because it is not within that value. If we cut out a side length of 3, we don't have enough material to do so.

So remember, 2 thirds is equal to about 0.67, which is easier to understand as a dimension rather than a fraction. So that is our critical point that we're working with here. Now that we have our critical points, we can move on to our final step here where we would either want to apply the extreme value theorem or the second derivative test. Here, since we do have endpoints, making this a closed interval, we can use the extreme value theorem to plug our critical points and our endpoints into our original function. Here, since we're trying to maximize the volume, we're going to be looking for the largest value to give us that max.

So plugging into our original function here, I'm going to go ahead and do that in this corner. We have our endpoints, 0 and 1.5. And we also have that fraction value. I'm just going to keep it as 2 thirds to avoid rounding errors. And we're plugging this into our original function, this polynomial 4x3−22x2+24x.

It turns out that for both of our endpoints, plugging in these values gives us a value of 0, which, again, commonly happens when working with a closed interval, but it's always a good idea to just double check. Then plugging in that 2 thirds, so that's 4×2333−22×2322+24×23 running out of room here. But if we go ahead and put this down here, this ends up giving us a volume value here of about 7.41. Now based on what we were working with above, we know that this is in cubic feet because it is a volume value. So 7.41 cubic feet.

This is clearly the largest value of the 3. So this represents our maximum volume. Now we were asked to find this maximum volume, so we would want to report that in our answer. Now we also wanted to tell here the dimensions of these little squares being cut out. And here, we already found that that dimension was 2 thirds or 0.67.

So the x value that creates this 7.41 cubic feet volume is 0.67 feet being cut out of those corners. So we have our final answer here with our maximum volume and the dimensions of those squares. Feel free to let me know if you have any questions here, and let's keep practicing.

Example 7: Shortest Rope

Here, we're going to work through another common type of applied optimization problem where we're looking to find the shortest length of rope, specifically a rope tied to the top of something down to the ground. Another variation of this problem that you may encounter is having a ladder leaning up against the house and trying to find the shortest possible ladder that you could use for that purpose. This is exactly the same problem and would follow the same process. So let's get into our problem here. We're told that we have a 10-meter tall tree standing in front of a rock cliff with its base 5 meters away from the base of the cliff.

We want to find here the length of the shortest rope that can be tied from the ground over the top of the tree to the cliff, just like we see in this image here. The first thing we want to do is draw a diagram and identify our variables here. So with this diagram that we already have, let's go ahead and label all the information given to us in the problem. We know that the height of this tree is 10 meters. We also know that this tree is 5 meters away from the base of the cliff here.

Now looking at this image, thinking about what variables I might want, I know that I'm going to be looking for the length of the shortest rope. So my rope should definitely get a variable. I'm going to refer to it as \( l \). Now looking at what's going on in this problem, seeing as I have some triangles forming here, I likely want my other side lengths of triangles to have variables as well. So I'm going to refer to the height of my cliff as \( h \) and this little missing piece of information here as \( x \).

So with those variables identified, let's move on to step 2 and write a function for our value being optimized. Now here, we want to find the length of the shortest rope. That means that we want to minimize \( l \). So we want to write a function representing \( l \). What is \( l \) equal to?

Well, thinking about what's going on here in my image, I can see that I have this triangle formed here. So that full side length \( l \) is the hypotenuse of my triangle, meaning that \( l^2 \) is equal to \( h^2 \), one side length, plus my other side length squared, which is this 5 added together with this \( x \) to make up that entire side length. This becomes \( (x+5)^2 \). Now I can go ahead and just take the square root of both sides to get \( l \) because I know that I wouldn't want the negative square root of this. It's totally fine to just go ahead and take that positive square root.

So looking at this, we want it in terms of just one variable. It is clearly not in terms of one variable because I have both \( h \) and \( x \). So we want to think about getting \( h \) in terms of \( x \) using the constraints that are put on our problem. Remember that it often requires an extra equation to do this. So let's think about how we could do that.

Here, I have 2 triangles that form. These triangles are similar. Because of this fact, I can use the ratio of my side lengths in order to express \( h \) in terms of \( x \). So I know that this height and this height, the ratios of that, so I can say \( \frac{h}{10} \) is equal to the ratio of the side lengths here.

So \( \frac{x+5}{x} \). Now from here, we can multiply both sides by 10 to get \( h \) by itself. That 10 will cancel, and I have that \( h = \frac{10(x+5)}{x} \). Now I can simplify this a little bit more if I just want to go ahead and cancel that \( x \) out and put the \( x \) with the 50 so that I don't have a denominator here.

So this gives me \( 10 + \frac{50}{x} \). So that's \( h \). And now I can plug \( h \) back into my equation for \( l \) to get it in terms of just one variable. So this is \( 10 + \frac{50}{x} \). All of that is squared plus \( x + 5 \) squared.

So with step 2 done, we move on to step 3, any domain restrictions. I know that \( x \) has to be greater than 0, but I don't really know anything other than that. I don't know how this length is going to work out. So I'm just going to move on to step 4 and go ahead and find my critical points where my first derivative is equal to 0. So we need to find our first derivative.

It may get a little bit messy just because we're working with a square root here. So \( l' \) is going to be equal to one-half times all of this stuff on the inside. So that's \( 10 + \frac{50}{x} \). That is squared, plus \( x \) o plus 5 squared. And all of that is to the power of negative one-half.

I'm using the chain rule here. Now I want to multiply by the derivative of everything inside, and I'm going to have to use the chain rule again. This ends up being 2 times \( 10 + \frac{50}{x} \) and then multiply it again. We're kind of using the chain rule multiple times over and over again. Multiply by the derivative of what's inside of these parentheses.

That's just going to be negative \( \frac{50}{x^2} \). I can't forget that I also have this term here. This is plus 2 times \( x + 5 \) squared or just 2 times \( x+5 \), not squared. And then that, I don't have to use a chain rule because the derivative of the inside is just 1. So this is a little bit messy, but let's clean it up a little bit by just moving some stuff around here.

This one half and all of these twos will cancel because they are all multiplying each other. This negative one-half means that this is on the bottom of a fraction here. So I'm going to put all of this on the top of my fraction, all of this on the bottom. So this is \( 10 + \frac{50}{x} \) times negative \( \frac{50}{x^2} \) plus \( x + 5 \), all divided by the square root since that's to the power of negative one-half of \( 10 + \frac{50}{x} \) squared plus \( x + 5 \) squared. Okay.

So this is our derivative here. Now we're tasked with finding where this derivative is equal to 0, which may seem like it's going to be really complicated, but let's walk through this together. Now when we're trying to find where our first derivative is equal to 0, we can really just take the numerator and set that equal to 0. So this denominator doesn't really matter to me right now. I just want to work with the numerator.

But in working with the numerator, I need to do a lot of simplification in order to actually figure out what \( x \) is. So working with my numerator here, I'm going to go ahead and distribute that negative \( \frac{50}{x^2} \) into my parentheses here. This gives me negative \( \frac{500}{x^2} \) plus negative or minus, minus \( \frac{25100}{x^3} \) and then plus \( x + 5 \). Now we're setting all of this equal to 0. Now this may look like we have nowhere to go, but I think a good step here would be to multiply everything by \( x^3 \).

So if we multiply absolutely everything in this function by \( x^3 \), let's see what happens. This gives us negative 500x minus 25100 plus \( x \) to the power of 4 plus 5\( x^3 \). And all of that is equal to 0. Now at this point, it's always a good idea to think about factoring. I'm going to go ahead and rewrite this in standard form with those descending powers.

So \( x^4 + 5x^3 - 500x - 25100 \). Now looking at this, it may not be immediately clear, but we can actually factor this by grouping. So I'm going to think about factoring out of these two terms and of these two terms. So let's go ahead and factor an \( x^3 \) out of those first two terms. This gives us \( x^3 \times (x + 5) \).

Then if I go ahead and factor a negative 500 out of both of these terms, that seems like a good choice, negative 500, this becomes \( x + 5 \). Now when this happens, we can see that this becomes \( (x^3 - 500) \times (x + 5) = 0 \). Now from this point, we can take each of these individual factors and set them equal to 0. So we've finally reached that point. I'm going to go ahead and restart up here just so I have a little bit more space.

So taking \( x^3 - 500 \)