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All courses Geometry · G-MG.3 54 of 57
Use geometric methods to solve design problems under constraints such as cost, space, or ratios.

Model and maximize fixed-fencing rectangle area

Problem
A rectangle has perimeter \(40~\text{units}\), and \(\text{one}~\text{side}\) is \(x\). Use the diagram to build a \(\text{one}\text{-}\text{variable}\) area model with its positive feasible domain, then find the maximizing dimensions and maximum area.
Prompt rectangle with sides x and y and fixed perimeter 40, so 2x+2y=40; the area model, feasible domain, optimizer, dimensions, and maximum area are withheld. Open full size
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Hint

Draw and label how many sides have length x and how many have length y.

Translate fencing into a linear constraint, solve for y, substitute into A=xy, restrict both dimensions positive, then find the quadratic vertex.

Solution walkthrough

01

Use the perimeter constraint

\[2x+2y=40~->~x+y=20~->~y=20-x\]

The diagram labels the rectangle sides x and y. Solving the fixed-perimeter equation expresses y using x alone.

02

Build the one-variable area model

\[A=x*y~->~A(x)=x(20-x)=-x^2+20x\]

Substitute y=20-x into length times width.

03

State the positive feasible domain

\[x>0~\text{and}~20-x>0~->~0<x<20\]

Both side lengths must be positive; the endpoints would collapse the rectangle to zero area.

04

Find the quadratic maximum

\[A(x)=-(x-10)^2+100\]

Completing the square shows a downward-opening parabola whose vertex occurs at x=10 with area 100.

05

Recover dimensions and conclude

\[x=10;~y=20-10=10.~\text{From}~2x~+~2y~=~40,~y~=~20~-~x~\text{and}~A(x)~=~x(20~-~x),~\text{with}~0~<~x~<~20.~\text{The}~\text{maximum}~\text{occurs}~\text{at}~x~=~10,~\text{giving}~a~10-by-10~\text{rectangle}~\text{with}~\text{area}~100~\text{square}~\text{units}.\]

The dimensions satisfy the perimeter check 2(10)+2(10)=40 and give the model's maximum.

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Another way

  1. Use the vertex formula for A=-x²+20x: x=-b/(2a)=-20/(-2)=10, then find y and area.

!

Common mistake

Do not optimize x(40-x). The perimeter counts each side twice, so x+y=20 and the other side is 20-x.

Answer rectangle with 2x+2y=40, so y=20-x and A=x(20-x) on 0<x<20; the downward parabola has vertex x=10, giving dimensions 10 by 10 and maximum area 100 square units.

Solution walkthrough video