Maximum-area rectangles Of all rectangles with a fixed perimet...

Question

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

Accepted Answer

Welcome back, everyone. An artist is creating a rectangular mosaic with a total boundary length of centimeters to ensure the maximum possible area for his artwork. What should the dimensions of the mosaic be? Write the dimensions in terms of P were given for answer choices A P. By 2 multiplied by divided by 2, B divided by 4 by divided by 4, C P divided by 3 by P divided by 3, and D P divided by 5 by P divided by 5, all given in centimeters by centimeters. So essentially what we have to do is visualize a rectangle. We know that it has 2 lengths and 2 widths. And specifically we can say that the perimeter of such a rectangle is 2 lengths plus 2 width. Now if we divide both sides by 2, we're going to get length plus width is equal to perimeter divided by 2. And what we can do is simply express width in terms of length. So width is equal to perimeter divided by 2 minus length. Now we want to Ensure the maximum possible area and we know that the area is equal to length multiplied by width. We're going to use land and we're going to express with in terms of land, we already have that, so we're multiplying land by. Perimeter divided by 2 minus length and if we expand that function we're going to get parameter divided by 2 multiplied by length minus length squared. Remember that perimeter is a constant. We can say that it is constant. And from here we want to maximize area, so we're going to find the derivative. So the derivative of area with respect to land. So let's differentiate. P divided by 2 L. Minus L2. So here we have to apply the Power rule For L, the derivative is 1, so we get P divided by 2, and the derivative of minus Lhere is -2 L. And now we want to set it equal to 0 to identify the critical points. So 2 L is equal to P divided by 2. Therefore, L is equal to p divided by 4. And then we want to make sure that this is a point of maximum. So let's apply the 2nd derivative test and let's identify the second derivative of the area function. So we're differentiating he divided by 2 minus 2 L. So the first term is a constant, its derivative is 0, and the derivative of L is 1, so we get -2, which is less than 0 for any value of L, including the critical point, right? So according to the second derivative test, we have a function that is concave down. In this Means that we have a local maximum within our function at the critical point. So we have verified that length equals p divided by 4 gives us a maximum. And this means that with, which is equal to p divided by 2 minus L, is going to be equal to p divided by 2 minus p divided by 4 because that's what L is at the maximum point. And if we perform the calculations using the common denominator of 4, we get 2 p minus p divided by 4, which is also P divided by 4. So the dimensions that ensure the maximum area are going to be equal to P divided by 4 centimeters by P divided by 4 centimeters for length and width respectively, which corresponds to the answer choice B. Thank you for watching.

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

Key Concepts

Perimeter and Area of Rectangles

Optimization in Calculus

Critical Points and the First Derivative Test

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