Question

a rectangle is constructed with its base on the x - axis and two of its vertices above the x - axis on the parabola ( y = 64 - x^{2} ). what are the dimensions of the rectangle with the maximum area? what the area?
in the rectangle with the maximum area, the shorter dimension is about 9.24 and the longer dimension is about 42.67
(round to two decimal places as needed.)
the maximum area of the rectangle is about (square)
(round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the area of a rectangle

The area of a rectangle \(A = length\times width\). Let the \(x -\)coordinate of the right - hand vertex of the rectangle (above the \(x -\)axis) be \(x\). Then the length of the base of the rectangle (along the \(x -\)axis) is \(2x\) (since the rectangle is symmetric about the \(y -\)axis), and the height of the rectangle is \(y=64 - x^{2}\). So the area function \(A(x)=2x(64 - x^{2})=128x-2x^{3}\), where \(x>0\).

Step2: Find the derivative of the area function

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(A^\prime(x)=\frac{d}{dx}(128x-2x^{3})\).
\(A^\prime(x)=128-6x^{2}\).

Step3: Find the critical points

Set \(A^\prime(x) = 0\), so \(128-6x^{2}=0\).
Rearrange the equation: \(6x^{2}=128\), then \(x^{2}=\frac{128}{6}=\frac{64}{3}\), and \(x=\sqrt{\frac{64}{3}}\approx4.62\) (we take the positive value since \(x>0\)).

Step4: Find the second - derivative of the area function

Differentiate \(A^\prime(x)\) with respect to \(x\). \(A^{\prime\prime}(x)=\frac{d}{dx}(128 - 6x^{2})=-12x\).
When \(x = \sqrt{\frac{64}{3}}\), \(A^{\prime\prime}(\sqrt{\frac{64}{3}})=- 12\sqrt{\frac{64}{3}}<0\), so the function \(A(x)\) has a maximum at \(x=\sqrt{\frac{64}{3}}\).

Step5: Calculate the area

Substitute \(x=\sqrt{\frac{64}{3}}\) into the area function \(A(x)\).
\(A(\sqrt{\frac{64}{3}})=128\sqrt{\frac{64}{3}}-2(\sqrt{\frac{64}{3}})^{3}\)
\(A(\sqrt{\frac{64}{3}})=128\sqrt{\frac{64}{3}}-2\times\frac{64}{3}\sqrt{\frac{64}{3}}\)
\(A(\sqrt{\frac{64}{3}})=(128-\frac{128}{3})\sqrt{\frac{64}{3}}=\frac{256}{3}\sqrt{\frac{64}{3}}\)
\(A(\sqrt{\frac{64}{3}})\approx\frac{256}{3}\times4.62\approx395.02\)

Answer:

The maximum area of the rectangle is about \(395.02\)