022 Exam 2 Sample A, Problem 10 - Revision history

MathAdmin at 17:46, 16 May 2015

2015-05-16T17:46:19Z

MathAdmin at 15:14, 16 May 2015

2015-05-16T15:14:14Z

MathAdmin: Created page with "350px '''Use calculus to set up and solve the word problem:''' Find the length and width of a rectangle that has a perimet..."

2015-05-16T04:37:44Z

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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[[File:022_2_A_10GP.png|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 perimeter of 48 meters and maximum area.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|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:
|-
|
::<math>P\,=\,2x+2y,</math>
|-
|and that for area:
|-
|
::<math>A\,=\,xy.</math>
|-
|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:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|'''Express one variable in terms of the other:''' Since we know that the perimeter is to be 48 meters and <math style="vertical-align: -20%">P\,=\,2x+2y</math>, we can solve for <math style="vertical-align: -15%">y</math> in terms of <math style="vertical-align: 0%">x</math>. Since
|-
|
::<math>48\,=\,P\,=\,2x+2y,</math>
|-
|we find that <math style="vertical-align: -20%">y=24-x.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|'''Find an expression for area in terms of one variable:''' Now, we can use the substitution from part 1 to find
|-
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::<math>A(x)\,=\,xy\,=\,x(24-x)\,=\,24x-x^2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|'''Find the derivative and its roots:''' Since <math style="vertical-align: -22%">A(x)=24x-x^2</math>, we have
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::<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.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|The area is maximized when both the length and width are 12 meters.
|}

[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|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:
+|As with all geometric word problems, it helps to 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:
 |-
+|

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|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:
+|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:
 |-
+|
 ::<math>P\,=\,2x+2y,</math>
 |-
-|and that for area:
+|and for area:
 |-
+|
 ::<math>A\,=\,xy.</math>
 |-
 |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.
 |}
@@ Line 50: / Line 50: @@
 ::<math>A'(x)\,=\,24-2x\,=\,2(12-x).</math>
 |-
-|This derivative is zero precisely when <math style="vertical-align: -20%">x=y=12</math>, and these are the values that will maximize area.
+|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!
 |}