Maximum printable area A rectangular page in a text (with wi...

Question

Maximum printable area A rectangular page in a text (with width x and length y) has an area of 98 in² , top and bottom margins set at 1 in, and left and right margins set at 1/2 in. The printable area of the page is the rectangle that lies within the margins. What are the dimensions of the page that maximize the printable area?

Accepted Answer

A rectangular swimming pool is being designed with an area of 150 m square. Surrounding the pool on all sides is a uniform 2 m wide deck, but the dimensions of the pool that minimize the total area, including the deck. We have 4 possible answers, which are length and width dimensions of the pool and meters. Now, to solve this, let's look at what we currently know. We know that we have a pool. Where the area is equals to 150 m squared. Length times width will then be equal to 150 m square. Now, we have the total area. Which is the length total multiplied by the width total. Now, we notice that this is a rectangular pool. With a 2 m wide deck on all sides. This means then that there is 4 m on both sides of the rectangle, both in the length and the width direction. We can then write this. As L + 4 multiplied by W + 4. So now we want to maximize this area of the pool. Let's look at our first equation. We can create a substitution. By saying length multiplied by width equals 150, and dividing both sides by length, to get width equals 150 divided by length. We can then substitute this into our area total equation. L + 4 multiplied by 150 divided by L plus 4. Now we can simplify this expression by expanding. We can multiply these together to get L multiplied by 150 divided by L plus 4. Plus 4 multiplied by 150 divided by L plus 4. We can simplify. To get 150 Plus 4 L Plus 600 divided by L. Plus 16. Which gives us 166 plus 4L. Plus 600 divided by L. That is our total area equation. Now, to minimize the tolar area, we need to set our first derivative equal to zero. So let's find a prime. A is equals to 0. Plus 4 Minus 600 divided by L2d. This is our derivative. Which is going to be set equal to 0. 0 equals 4 minus 600 divided by L2. Now, we can solve this. By moving the 4 over, we subtract the 4. We get -4 equals -600 divided by L2. And if we keep solving, We get L2 equals 600 divided by 4. Or 150 L will then be the square root of 150. We can simplify this To give us 5 square of 6 in meters. Now, we can verify if this is a maximum by checking our second derivative. A double prime It equals So if we look at our function, we have 600 divided by L2. This derivative would be 1200. Divided by the 3rd. Now, we just need to make sure that this is greater than 0. And because L is in a denominator, as long as L is greater than 0, this derivative will be greater than 0 overall. If L is greater than 0, 2nd derivative is greater than 0. That would make this a minimum. Now that we've confirmed our minimum, we can find width. You know, width is equals to 150 divided by L, which is 150 divided by the square root. Of 150 We can simplify this. This will just give us the square root of 150 or 5 square roots of 6 m. So the answer with the dimensions of the pool that minimize this area. Is 5 square to 6 m by 52 to 6 m. If we look at our possible answers, we see the answers to this problem. His answer is C. OK, hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Optimization

Area of a Rectangle

Constraints

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