Question

a rectangle is inscribed with its base on the x-axis and its upper corners on the parabola $y = 8 - x^2$. what are the dimensions of such a rectangle with the greatest possible area?
width =
height =
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Explanation:

Step1: Set up the area function

Let the \(x -\)coordinate of the upper - right corner of the rectangle be \(x\). The width of the rectangle is \(w = 2x\) (since the rectangle is symmetric about the \(y -\)axis), and the height \(h=y = 8 - x^{2}\).
The area function \(A(x)\) of the rectangle is \(A(x)=(2x)(8 - x^{2})=16x-2x^{3}\), where \(x>0\).

Step2: Find the derivative of the area function

Differentiate \(A(x)\) with respect to \(x\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(A^\prime(x)=\frac{d}{dx}(16x-2x^{3})=16-6x^{2}\).

Step3: Find the critical points

Set \(A^\prime(x) = 0\), so \(16-6x^{2}=0\).
Rearrange the equation: \(6x^{2}=16\), then \(x^{2}=\frac{16}{6}=\frac{8}{3}\), and \(x=\sqrt{\frac{8}{3}}=\frac{2\sqrt{6}}{3}\) (we take \(x>0\) since \(x\) represents a length).

Step4: Use the second - derivative test

Differentiate \(A^\prime(x)\) to get the second - derivative. \(A^{\prime\prime}(x)=\frac{d}{dx}(16 - 6x^{2})=-12x\).
When \(x = \frac{2\sqrt{6}}{3}\), \(A^{\prime\prime}(\frac{2\sqrt{6}}{3})=-12\times\frac{2\sqrt{6}}{3}=-8\sqrt{6}<0\). So \(A(x)\) has a maximum at \(x=\frac{2\sqrt{6}}{3}\).

Step5: Calculate the width and height

The width \(w = 2x\), substituting \(x=\frac{2\sqrt{6}}{3}\), we get \(w=\frac{4\sqrt{6}}{3}\).
The height \(h=8 - x^{2}\), substituting \(x^{2}=\frac{8}{3}\), we get \(h=8-\frac{8}{3}=\frac{24 - 8}{3}=\frac{16}{3}\).

Answer:

Width \(=\frac{4\sqrt{6}}{3}\), Height \(=\frac{16}{3}\)