022 Exam 2 Sample A, Problem 10

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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 and as shown in the image, we need to remember the equations of a rectangle for perimeter:
and for area:
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 , we can solve for in terms of . Since
we find that
Step 2:  
Find an expression for area in terms of one variable: Now, we can use the substitution from part 1 to find
Step 3:  
Find the derivative and its roots: Since , we have
This derivative is zero precisely when , 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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