Maximum-area rectangles Of all rectangles with a perimeter of ...

Question

Maximum-area rectangles Of all rectangles with a perimeter of 10, which one has the maximum area? (Give the dimensions.)

Accepted Answer

Welcome back, everyone. A farmer wants to fence a rectangular area for her sheep using 200 m of fencing material. What are the dimensions of the rectangle that will maximize the area for her sheep? Were given for answer choices A says 150 m by 50 m. B says 25 m by 50 m. C 45 m by 50 m, and D 25 m by 25 m. So let's visualize a rectangle. Let's suppose that we have a length L and with W. So essentially we're given the perimeter, right? The perimeter is going to be equal to 200 m. This is a fixed value, and then we can express the perimeter as a sum of the side lens. So essentially we're going to take 2 multiplied by L plus 2 multiplied by W. This is our constraint that is equal to 200. And we can essentially divide both sides by 2, which gives us a constraint of L plus W equals 200. Now what we can do is simply express width in terms of length for width equals. 200 We're actually 100, right? Because we divided both sides by 2, and this means that the right hand side is now 100. So width is equal to 100 minus length. We're going to use this equation in our optimization. And what we want to optimize is the area function. Now area is equal to length multiplied by width, and we can see that this is simply a length multiplied by 100 minus length from the first equation. Now if we distribute the terms, our area function is 100 l minus 12, and now we want to optimize it so we can differentiate area. With respect to length, the a divided by DL is simply the derivative of 100 l minus L2. Applying the power rule, we're going to get 100 minus 2 L. And we want to set that equal to 0 to identify the critical points. So this means that 2 L is equal to 100. And thereforel is equal to 50. Now with It's going to be equal to 100 minus 50, which also gives us 50. But now we want to check if we have a point of maximum, and we're going to use the second derivative test. So we're going to find the second derivative of area. With respect to land. And this means that we're differentiating 100 minus 2 L, the derivative of that is -2, which is negative for every L, including the critical point. The 2nd derivative test tells us that the function is concave down if the 2nd derivative is negative, so we indeed have a local maximum. Which verifies that these dimensions correspond to. The maximum area. So we have 50 m by 50 m, which corresponds to the answer to A. Thank you for watching.

Maximum-area rectangles Of all rectangles with a perimeter of 10, which one has the maximum area? (Give the dimensions.)

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