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 $x$ and $y$ as shown in the image, we need to remember the equations for perimeter:

 $P\,=\,2x+2y,$

and that for area:

 $A\,=\,xy.$

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 $P\,=\,2x+2y$, we can solve for $y$ in terms of $x$. Since

 $48\,=\,P\,=\,2x+2y,$

we find that $y=24x.$

Step 2:

Find an expression for area in terms of one variable: Now, we can use the substitution from part 1 to find

 $A(x)\,=\,xy\,=\,x(24x)\,=\,24xx^{2}.$

Step 3:

Find the derivative and its roots: Since $A(x)=24xx^{2}$, we have

 $A'(x)\,=\,242x\,=\,2(12x).$

This derivative is zero precisely when $x=y=12$, and these are the values that will maximize area.

Final Answer:

The area is maximized when both the length and width are 12 meters.

Return to Sample Exam