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 geometric word problems, it helps to start with a picture. Using the variables $x$ and $y$ as shown in the image, we need to remember the equations of a rectangle for perimeter:
$P\,=\,2x+2y,$
and 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 $y=24-x.$

Step 2:
Find an expression for area in terms of one variable: Now, we can use the substitution from part 1 to find
$A(x)\,=\,xy\,=\,x(24-x)\,=\,24x-x^{2}.$

Step 3:
Find the derivative and its roots: Since $A(x)=24x-x^{2}$, we have
$A'(x)\,=\,24-2x\,=\,2(12-x).$
This derivative is zero precisely when $x=y=12$, and these are the values that will maximize area. Also, don't forget the units - meters!

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

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