Difference between revisions of "022 Exam 2 Sample A, Problem 10"

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(Created page with "right|350px <span class="exam"> '''Use calculus to set up and solve the word problem:''' Find the length and width of a rectangle that has a perimet...")
 
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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|As with all word problems, start with a picture.  Using the variables <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">y</math> as shown in the image, we need to remember the equations for perimeter:
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|As with all word problems, start with a picture.  Using the variables <math style="vertical-align: 0%">x</math> and <math style="vertical-align: -20%">y</math> as shown in the image, we need to remember the equations of a rectangle for perimeter:
 
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::<math>P\,=\,2x+2y,</math>
 
::<math>P\,=\,2x+2y,</math>
 
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|and that for area:
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|and for area:
 
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::<math>A\,=\,xy.</math>  
 
::<math>A\,=\,xy.</math>  
 
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|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.  
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|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.
 
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::<math>A'(x)\,=\,24-2x\,=\,2(12-x).</math>
 
::<math>A'(x)\,=\,24-2x\,=\,2(12-x).</math>
 
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|This derivative is zero precisely when <math style="vertical-align: -20%">x=y=12</math>, and these are the values that will maximize area.
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|This derivative is zero precisely when <math style="vertical-align: -20%">x=y=12</math>, and these are the values that will maximize area. Also, don't forget the units - meters!
 
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Revision as of 08:14, 16 May 2015

022 2 A 10GP.png

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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