# 022 Exam 2 Sample A, Problem 10

Use calculus to set up and solve the word problem: Find the length and width of a rectangle that has a perimeter of 48 meters and maximum area.

Foundations:
As with all word problems, start with a picture. Using the variables ${\displaystyle x}$ and ${\displaystyle y}$ as shown in the image, we need to remember the equations of a rectangle for perimeter:
${\displaystyle P\,=\,2x+2y,}$
and for area:
${\displaystyle A\,=\,xy.}$
Since we want to maximize area, we will have to rewrite it as a function of a single variable, and then find when the first derivative is zero. From this, we will find the dimensions which provide the maximum area.

Solution:

Step 1:
Express one variable in terms of the other: Since we know that the perimeter is to be 48 meters and ${\displaystyle P\,=\,2x+2y}$, we can solve for ${\displaystyle y}$ in terms of ${\displaystyle x}$. Since
${\displaystyle 48\,=\,P\,=\,2x+2y,}$
we find that ${\displaystyle y=24-x.}$
Step 2:
Find an expression for area in terms of one variable: Now, we can use the substitution from part 1 to find
${\displaystyle A(x)\,=\,xy\,=\,x(24-x)\,=\,24x-x^{2}.}$
Step 3:
Find the derivative and its roots: Since ${\displaystyle A(x)=24x-x^{2}}$, we have
${\displaystyle A'(x)\,=\,24-2x\,=\,2(12-x).}$
This derivative is zero precisely when ${\displaystyle x=y=12}$, and these are the values that will maximize area. Also, don't forget the units - meters!