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Chapter 16: Problem 373

A piece of wire 20 inches long is to be cut and made into a rectangular frame. What dimensions should be chosen so that the area of the rectangle enclosed is maximal?

Short Answer

Expert verified
The dimensions of the rectangle that would maximize its area are length x = 5 inches and width y = 5 inches.

Step by step solution

01

Set Up the Variables

Let x be the length of the rectangle and y be the width. Since the wire is 20 inches long and it forms the entire perimeter of the rectangle, we have the following equation: P = 2x + 2y = 20 We need to find the dimensions x and y, such that the area of the rectangle is maximized. Area (A) = xy
02

Write the Constraint Equation

First, we need to write the constraint equation for the perimeter of the rectangle. As stated earlier, the wire forms the entire perimeter of the rectangle, so the equation can be written as: 2x + 2y = 20 Now, we need to solve for y in terms of x: y = 10 - x
03

Rewrite the Area Equation

Now we will rewrite the area equation in terms of x by substituting our constraint equation (y = 10 - x) into the area equation: A(x) = x(10 - x) A(x) = 10x - x^2
04

Find the Critical Points

In order to find the maximum area, we need to find the critical points of the area function A(x) by taking the derivative with respect to x and setting it equal to 0. A'(x) = 10 - 2x Now, set A'(x) to 0 to find the critical points: 10 - 2x = 0 2x = 10 x = 5
05

Calculate Dimensions for Maximum Area

With the critical point x = 5, substitute it back into the constraint equation (y = 10 - x) to find the corresponding y value: y = 10 - 5 = 5 Therefore, the dimensions of the rectangle that would maximize its area are length x = 5 inches and width y = 5 inches.

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