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Chapter 5: Problem 50

Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

Short Answer

Expert verified
The length and width of the rectangle are 8 feet and 12 feet, or vice versa.

Step by step solution

01

Set up the equations

Use the given perimeter and area to set up two equations. We know that the perimeter \( P=40 \) which gives the equation \( 2l+2w=40 \) and the area \( A=96 \) which gives the equation \( lw=96 \).
02

Solve for one variable

From the perimeter equation, solve for one variable, for example \( w \), which gives \( w=20-l \).
03

Substitute and solve

Substitute \( w=20-l \) into the area equation, to get \( l(20-l)=96 \). By simplifying, this results into a quadratic equation \( l^2-20l+96=0 \). Solving this quadratic equation yields two solutions \( l=8 \) and \( l=12 \).
04

Find the corresponding width

Substitute the length into \( w=20-l \) to get the corresponding width. For \( l=8 \), we find \( w=12 \) and for \( l=12 \), we get \( w=8 \).

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