Question

section 3.3
score: 0/12 answered: 1/12
question 12
a rectangle is inscribed with its base on the x - axis and its upper corners on the parabola ( y = 6 - 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 right - hand upper corner of the rectangle be \(x\). Then the width of the rectangle is \(w = 2x\) (since the rectangle is symmetric about the \(y -\)axis), and the height of the rectangle is \(h=y = 6 - x^{2}\).
The area function \(A(x)\) is given by \(A(x)=w\times h=(2x)(6 - x^{2})=12x-2x^{3}\), where \(x>0\).

Step2: Find the derivative of the area function

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative of \(A(x)\) is \(A^\prime(x)=\frac{d}{dx}(12x - 2x^{3})\).
\(A^\prime(x)=12-6x^{2}\).

Step3: Find the critical points

Set \(A^\prime(x) = 0\), so \(12-6x^{2}=0\).
Rearrange the equation: \(6x^{2}=12\), then \(x^{2} = 2\), and \(x=\sqrt{2}\) (since \(x>0\)).

Step4: Check the second - derivative

Find the second - derivative \(A^{\prime\prime}(x)=\frac{d}{dx}(12 - 6x^{2})=-12x\).
When \(x = \sqrt{2}\), \(A^{\prime\prime}(\sqrt{2})=-12\sqrt{2}<0\). So the function \(A(x)\) has a maximum at \(x=\sqrt{2}\).

Step5: Calculate the width and height

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

Answer:

Width \(=2\sqrt{2}\), Height \(=4\)