/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A rectangle is bounded by the \(... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 25

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

Expert verified
The rectangle has maximum area when its length and width are \(5\sqrt{2}\) and \(\frac{5}{\sqrt{2}}\) respectively.

Step by step solution

01

Express the area function in terms of \(x\)

The area \(A\) of the rectangle can be expressed as the product of its length and its width. The length of the rectangle is \(2x\), and its width, bounded by the semicircle, can be described as \(y = \sqrt{25 - x^2}\), therefore the area \(A\) in terms of \(x\) can be expressed as \(A(x) = 2x.(\sqrt{25 - x^2})\)
02

Calculate the derivative of the area function

The next step is to find the derivative of the area function to find the critical points. The derivative \(A'(x)\) is given by the formula \[A'(x) = 2(\sqrt{25 - x^2}) + 2x.\frac{-x}{\sqrt{25 - x^2}} = 2\sqrt{25-x^2} - 2x^2/(\sqrt{25-x^2})\].
03

Find critical points

By setting the derivative equal to zero, we can solve for \(x\) to find the critical points: \(2\sqrt{25 - x^2} - 2x^2/(\sqrt{25 - x^2}) = 0\). One solution to this is \(x = \frac{5}{\sqrt{2}}\).
04

Confirm the maximum

Substituting the critical value \(x = \frac{5}{\sqrt{2}}\) into the second derivative of \(A(x)\) (namely \(A''(x)\)) will determine if this point is a maximum. A negative result confirms a maximum point. \(A''(x)\) is found to be negative when \(x = \frac{5}{\sqrt{2}}\).
05

Find length and width

Finally, we can substitute \(x = \frac{5}{\sqrt{2}}\) back into our original bound (i.e., \(y = \sqrt{25 - x^2}\)) to find \(y = \frac{5}{\sqrt{2}}\). Hence, the rectangle has maximum area when its length is \(2x = 2 \cdot \frac{5}{\sqrt{2}} = 5\sqrt{2}\) and its width is \(y = \frac{5}{\sqrt{2}}\).

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