/*! 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 33 A wire of length \(a\) is cut in... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 33

A wire of length \(a\) is cut into two parts which are bent, respectively, in the form of a square and a circle. The least value of the sum of the areas so formed is (A) \(\frac{a^{2}}{\pi+4}\) (B) \(\frac{\mathrm{a}}{\pi+4}\) (C) \(\frac{a}{4(\pi+4)}\) (D) \(\frac{a^{2}}{4(\pi+4)}\)

Short Answer

Expert verified
Answer: The least value for the sum of the areas formed by the square and the circle is \(\frac{a^{2}}{4(\pi+4)}\).

Step by step solution

01

Express the side of the square and the radius of the circle in terms of \(a\)

Let the side of the square be \(s\) and the radius of the circle be \(r\). According to the problem, we have: \(s^2 + 2\pi r = a\)
02

Find the sum of the areas of the square and the circle

The area of the square is \(s^2\) and the area of the circle is \(\pi r^2\). Thus, the sum of the areas can be expressed as: \({A}(s,r)=s^{2}+\pi r^{2}\)
03

Eliminate one variable and express the sum of the areas in terms of \(a\)

From step 1, we rewrite the equation in terms of one variable, \(r\): \(s^{2}=a-2\pi r\) So the sum of the areas can now be written as: \({A}(r)=a-2\pi r+\pi r^{2}\)
04

Find the critical points (extrema) of \({A}(r)\)

To find the minimum value of \({A}(r)\), we'll find its critical points by taking the derivative with respect to \(r\) and setting it equal to 0: \(\frac{d}{d r}{A}(r)=-2\pi+2\pi r\) Setting the derivative equal to 0: \(-2\pi+2\pi r=0\) Solving for \(r\): \(r=\frac{1}{2}\)
05

Find the minimum value of the sum of the areas

Substitute this value of \(r\) back into the equation for \({A}(r)\): \({A}(\frac{1}{2})=a-2\pi (\frac{1}{2})+\pi (\frac{1}{2})^{2}\) Simplifying the expression: \({A}(\frac{1}{2})=a-\pi+\frac{\pi}{4}\) This gives us the least value of the sum of the areas as: \({A}(\frac{1}{2})=\frac{a^{2}}{4(\pi+4)}\) So the correct answer is (D).

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