Problem 50

Question

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

Step-by-Step Solution

Verified
Answer
The length and width of the rectangle are 8 feet and 12 feet, or vice versa.
1Step 1: 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 \( lw=96 \).
2Step 2: Solve for one variable
From the perimeter equation, solve for one variable, for example \( w \), which gives \( w=20-l \).
3Step 3: 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 \( l=12 \).
4Step 4: 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 \( w=8 \).