Your company can manufacture x hundred grade A tires and y hun...

Question

Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where 0 ≤ x ≤ 4 and y = (40 - 10x)/(5-x). Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says a bread factory can produce 100 loaves of whole wheat bread and 100 loaves of rye bread a day, where 0 is less than or equal to X, which is less than or equal to 3, and Y is equal to the quantity of 45 minus 15 X in quantity, divided by the quantity of 5 minus X. The profit on a loaf of whole wheat bread is twice the profit on a loaf of rye bread. To maximize profit, which should be the quantity of each type to produce. We have 4 possible choices as our answers. For choice A, we have about 256 loaves of whole wheat bread and 716 loaves of rye bread. For choice B, we have about 113 loaves of whole wheat bread and 725 loaves of rye bread. For choice C, about 256 loaves of whole wheat bread, and 633 loaves of rye bread. And for choice D, we have about 113 loaves of whole wheat bread and 611 loaves of rye bread. Now we're asked to determine how much of each type of bread needs to be produced in order to maximize profit. So, we're going to come up with a profit equation. We'll call that profit equation P, and we're gonna make that a function of X, which is going to be the 100 loaves of whole wheat bread, and this is going to be equal to 2 X. Plus why. Since we have twice the profit for the whole wheat bread, as we do for the rye bread, that's where the factor of 2 comes in, and the total profit is going to be the sum from each of those two types of loaves. Now we can actually write Y in terms of X, and we were given that expression in the problem. In doing so, this is going to be equal to 2 X. Plus the quantity of 45 minus 15 X in quantity, divided by the quantity of 5 minus X. So what we want to do now is actually maximize this function. And so the first thing we need to do is take a first derivative and find the critical points. So taking the derivative of our function P with respect to X will have P of X, and that's going to be equal to the derivative with respect to X. Of the quantity of 2 X. Plus, The quantity of 45 minus 15 X in quantity, divided by the quantity of 5 minus X. In quantity And so taking this derivative, this is going to be equal to. For the first term, we're gonna have the derivative with respect to X of 2 X, which is 2. Then we'll have plus, when we take the derivative, for the second term, we need to apply our quotient rule. So we call for the quotient world that we're going to have our denominator, which is going to be the quantity of 5 minus X in quantity, and that gets multiplied by the derivative of our numerators. So we'll have the derivative with respect to X of the quantity of 45 minus 15 X in quantity. And we take that derivative, we get -15. Then we'll have minus our numerator, which is the quantity of 45 minus 15 X in quantity, and that gets multiplied by the derivative of our denominator. So I take the derivative X of the quantity of 5 minus X, and when I take the derivative, I get minus 1. And both of those terms are actually divided by our denominator squared, so we'll have the quantity of 5 minus X in quantity squared. So simplifying our fraction in the second term here, this is going to be equal to 2. Plus The quantity will have -75. Plus 15 X. Plus 45 Minus 15 X. In quantity, divided by The quantity of 5 minus X in quantity squared. So simplifying further, this is going to be equal to 2 minus. 30 divided by the quantity of 5 minus X. In quantity squared. So here we've calculated our first derivative, and we need to identify critical points. So recall that critical points are the values of X at which our first derivative is undefined. Now, if we look here in the second term, we have a fraction, and our denominator in that fraction is equal to 0 when X is equal to 5. However, 5 is outside of our domain for X, so X has to be between 0 and 3. So, we're not going to look at that critical point. And so the other condition to have critical points is when our first derivative is equal to 0. So we're gonna set 2 minus 30 divided by the quantity of 5 minus X in quantity squared equal to 0. And we're going to solve this for X. So to solve this for X, the first step we'll do is add 30 divided by the quantity of 5 minus X in quantity squared to both sides. We'll have 2 is equal to 30 divided by the quantity of 5 minus X in quantity squared. So now I'll multiply both sides of the equation by the quantity of 5 minus X in quantity squared and divide both sides by 2. Doing so, we'll have quantity of 5 minus X. In quantity squared is equal to 15. So now I'll take the square root of both sides. In doing so, I'll have 5 minus X is equal to the square root of 15. Now, here we've just taken the positive square root, and that's because the quantity of 5 minus X has to be a positive quantity. Since X has to be between 0 and 3, that means that 5 minus X is always going to be positive. So, if I were to take the negative root, that means that that quantity would have to be negative, which is not a solution that I'm looking for in the domain of X. So now I'll need I'll need to do is solve this equation for X. In doing so, I'll get X is equal to 5 minus square root of 15. So now we need to test to make sure that this is a maximum value. So this is a critical point, and so it could be possibly a local maximum, a local minimum, or neither. So, what we're going to do is calculate our second derivative. So that means I need to calculate. P double prime of X. Which is going to be equal to the derivative with respect to X. Of the quantity of 2 minus 30 divided by the quantity of 5 minus X in quantity squared in quantity. So now I just need to take this derivative, so this is going to be equal to. When I take the derivative of 2 with respect to X, I get 0. Then for the next term, I will have minus 30. Multiplied by the derivative with respect to X of the quantity of. 5 minus X in quantity raised to the -2 power. And so when I take that derivative, I'll get -2 multiplied by the quantity of 5 minus X in quantity raised to the -3 power, and then still have to apply our chain rule, so we need to multiply this by derivative with respect X of the quantity of 5 minus X, which is -1. So simplifying this expression, this is going to be equal to minus 60. Divided by The quantity of 5 minus X in quantity duped. So now what I need to do is determine whether my second derivative here is positive or negative at my critical point. So we're going to calculate the double prime. of 5 minus the square root of 15. And this is going to be equal to. Minus 60 divided by. The quantity Of 5 minus the quantity of 5 minus the root of 15 in quantity. And all that is cubed in the denominator. And when simplify this expression, this is going to be equal to -60. Divided by the quantity of the square root of 15 in quantity, which is a negative value, so this is less than 0. And we call that if our second derivative is negative, that means that our function is concave down, which means that we do have a maximum at X equal to 5 minus the square root of 15. So that that X value that we found is The value of X that maximizes our profit. So now all we need to do is determine the number of lobes that we have of each type. So for this, we'll need to calculate our value for Y. We'll have Y is equal to, and here we'll substitute in for X, the 5 minus the square root of 15. So I'm gonna have Y is equal to the quantity of 45 minus 15 multiplied by the quantity of 5 minus the square roots of 15 in quantity. And all that is divided by the quantity. Of 5 minus quantity of 5 minus the square root of 15 in quantity. So simplifying this expression, this is going to be equal to. The square root of 15 minus 30, and that quantity is divided by. The square root of 15. And so now we need to do is actually calculate numerical answers for X and Y. So when we use our calculator to evaluate X, which is equal to 5 minus the square root 15, we get approximately equal to 1.13, and that is 1.1300 loaves of whole wheat bread. So we need to multiply that value by 100 to get the actual number of loaves. That means that this is going to be approximately equal to 113 loaves of whole wheat bread. That would do the same thing for our Y value. So we evaluate Y is equal to the quantity of the square root of 15 by 30 divided by the square root of 15. That is approximately equal to 7.25. And again, we'll need to convert that in from 100s of loaves to just the number of loaves, so we'll multiply that by 100. And so that means that this is going to be approximately equal to 725 loaves of rye bread. So those two values are the answers that we were looking for. So now, if we compare our answers that we calculated to our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where 0 ≤ x ≤ 4 and y = (40 - 10x)/(5-x). Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make?

Key Concepts

Constraints and Feasible Region

Profit Function

Optimization Techniques

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