Question

find the dimensions of a rectangle with a perimeter of 200 feet that has the maximum area. the side lengths are feet. (use a comma to separate answers as needed.)

Explanation:

Step1: Define variables

Let the length of the rectangle be \(x\) and the width be \(y\). The perimeter formula is \(P = 2(x + y)\), and given \(P=200\), so \(2(x + y)=200\), which simplifies to \(x + y=100\), and \(y = 100 - x\).

Step2: Express the area function

The area formula of a rectangle is \(A=xy\). Substitute \(y = 100 - x\) into the area formula, we get \(A(x)=x(100 - x)=100x - x^{2}\).

Step3: Find the derivative of the area function

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(A^\prime(x)=(100x - x^{2})^\prime=100-2x\).

Step4: Find the critical points

Set \(A^\prime(x) = 0\), so \(100-2x = 0\). Solving for \(x\):

$$
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Step5: Check the second - derivative (to confirm it's a maximum)

The second - derivative \(A^{\prime\prime}(x)=(100 - 2x)^\prime=-2\lt0\). Since \(A^{\prime\prime}(x)\lt0\) when \(x = 50\), the function \(A(x)\) has a maximum at \(x = 50\).
When \(x = 50\), then \(y=100 - x=50\)

Answer:

\(50,50\)