Question

1. rectangles beneath a parabola a rectangle is constructed with its base on the x - axis and two of its vertices on the parabola y = 48 - x². what are the dimensions of the rectangle with the maximum area? what is the area?

Explanation:

Step1: Set up the area function

Let the \(x\) - coordinate of the right - hand vertex of the rectangle on the parabola be \(x\). The base of the rectangle is \(b = 2x\) (since the rectangle is symmetric about the \(y\) - axis) and the height \(h=y = 48 - x^{2}\).
The area function \(A(x)\) of the rectangle is \(A(x)=b\times h=(2x)(48 - x^{2})=96x-2x^{3}\), where \(x>0\).

Step2: Find the derivative of the area function

Differentiate \(A(x)\) with respect to \(x\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(A^\prime(x)=\frac{d}{dx}(96x-2x^{3})=96 - 6x^{2}\).

Step3: Find the critical points

Set \(A^\prime(x) = 0\), so \(96 - 6x^{2}=0\).
Rearrange the equation: \(6x^{2}=96\), then \(x^{2}=16\).
Since \(x>0\), we get \(x = 4\).

Step4: Check the second - derivative

Differentiate \(A^\prime(x)\) to get the second - derivative \(A^{\prime\prime}(x)=\frac{d}{dx}(96 - 6x^{2})=-12x\).
When \(x = 4\), \(A^{\prime\prime}(4)=-12\times4=-48<0\). So the function \(A(x)\) has a maximum at \(x = 4\).

Step5: Find the dimensions and area

When \(x = 4\), the base \(b = 2x=8\).
The height \(h=48 - x^{2}=48-16 = 32\).
The area \(A=(2x)(48 - x^{2})\), substituting \(x = 4\) gives \(A=8\times32 = 256\).

Answer:

The base of the rectangle is \(8\), the height is \(32\), and the maximum area is \(256\).