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Chapter 3: Problem 7

Find the length and width of a rectangle that has the given perimeter and a maximum area. Perimeter: 100 meters

Short Answer

Expert verified
The rectangle that has a maximum area for the given perimeter of 100 meters is a square with side length of 25 meters.

Step by step solution

01

Express the area in terms of the width

Let's denote the width of the rectangle as \(w\). Because the perimeter of a rectangle is given by \(2(w + l)\), where \(l\) stands for length, we can express \(l\) as \((100 - 2w)/2\) or \(50 - w\). Now, since area equals width times length, it can be expressed as a function of \(w\): \(A = w(50 - w)\), which, when expanded gives \(A = 50w - w^2\).
02

Find the derivative of the area function

Next, we need to take the derivative of the area function and set it to zero to find the point where the area is maximized. The derivative of \(50w - w^2\) is \(50 - 2w\).
03

Solve equation for width

Setting the derivative equal to zero gives: \(50 - 2w = 0\), which solved for \(w\) gives \(w = 25\) meters.
04

Calculate the length of the rectangle

Substituting \(w = 25\) meters back into the equation for length gives \(l = 50 - 25 = 25\) meters.
05

Validation

From step 3 and 4 we notice that the rectangle is a square (as the width equals the length), which is a known result, that for given perimeter, a rectangle maximizes its area, when it is a square. Thus, our solution appears valid.

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