Difference between revisions of "022 Exam 2 Sample A, Problem 10"
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!Foundations:  !Foundations:  
    
−  As with all word problems, start with a picture. Using the variables <math style="verticalalign: 0%">x</math> and <math style="verticalalign: 20%">y</math> as shown in the image, we need to remember the equations of a rectangle for perimeter:  +  As with all geometric word problems, it helps to start with a picture. Using the variables <math style="verticalalign: 0%">x</math> and <math style="verticalalign: 20%">y</math> as shown in the image, we need to remember the equations of a rectangle for perimeter: 
    
   
Latest revision as of 10:46, 16 May 2015
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 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. 