Question

A rectangle is bounded by the \(x\) -axis and the semicircle \(y=\sqrt{25-x^{2}}\) (see figure). What length and width should the rectangle have so that its area is a maximum?

Short answer

The rectangle that maximizes the area under the semicircle \(y=\sqrt{25-x^{2}}\) has width \(2*\sqrt{25/2}\) and height \(\sqrt{25-25/2}\).

Step-by-step solution

Define the Area of Rectangle

The base of the rectangle ranges from -x to x, so the width (w) is 2x. The height (h) of the rectangle is given by \(y=\sqrt{25-x^{2}}\). The area A of the rectangle is therefore \(A = w * h = 2x * \sqrt{25-x^{2}}\).

Differentiate the Area with Respect to \(x\)

To find the maximum area, we need to find the point where the derivative of the area with respect to x is equal to zero. This can be done by applying the product rule and chain rule. The derivative of \(A\) with respect to \(x\) is \(A' = 2* \sqrt{25-x^{2}} - 2x^{2} / \sqrt{25-x^{2}}\).

Find Values of \(x\) for Which the Derivative is Zero

To find the critical points, set \(A'\) equal to zero and solve for \(x\). Simplify the equation to find that \(x = \sqrt{25/2}\).

Verify the Maximum Value

Verify that this value of \(x\) indeed maximizes the area by using the second derivative test. The second derivative of \(A\) with respect to \(x\) is negative, indicating that the area is maximized at \(x = \sqrt{25/2}\).

Find the Dimensions of the Rectangle

Substitute \(x = \sqrt{25/2}\) into the expressions for \(w\) and \(h\) to find the dimensions of the rectangle that maximizes its area. The width is \(w = 2x = 2*\sqrt{25/2}\) and the height is \(h = \sqrt{25-x^{2}} = \sqrt{25-25/2}\).