022 Sample Final A, Problem 2 - Revision history

MathAdmin at 23:32, 5 June 2015

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MathAdmin at 23:32, 5 June 2015

2015-06-05T23:32:14Z

MathAdmin at 23:29, 5 June 2015

2015-06-05T23:29:53Z

MathAdmin: Created page with "250px A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for his cows, the..."

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Created page with "right|250px <span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for his cows, the..."

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[[File:022_3_A_2GP.png|right|250px]]

<span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for his cows, the fenced pasture must contain 200 square meters of grass. If no fencing is required along the river, what dimensions will use the smallest amount of fencing?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|As a word problem, we must begin by assigning variables in order to construct useful equation(s). As an optimization problem, we will be taking a derivative of one of our equations in order to find an extreme point.
|}
 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|'''Declare Variables:'''   We are attempting to enclose a rectangular area, such that we use as little fencing as possible for the three fenced sides. Let's use <math style="vertical-align: 0%;">x</math> and <math style="vertical-align: -18%;">y</math> as indicated in the image, and simply call the length of fencing required <math style="vertical-align: 1%;">L</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|'''Form the Equations:'''   With the variables declared, we need two fences of length <math style="vertical-align: 0px">x</math> and one fence of length <math style="vertical-align: 0px;">y</math>. Together, we need a total length of <math style="vertical-align: -3px%;">L=2x+y</math>.
|-
|On the other hand, we know that the pasture has a fixed area of <math style="vertical-align: 0%;">200</math> square meters. Thus, we also know that <math style="vertical-align: -3px;">xy=200</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|'''Optimize:'''   Since <math style="vertical-align: -18%;">xy=500</math>, we also know <math style="vertical-align: -21%;">y=500/x</math>. Plugging this into our equation for length, we have
|-
|     <math>L(x) = 8x + 5\cdot \frac{500}{x} = 8x + \frac{2500}{x}.</math>
|-
|We now take the derivative to find
|-
|     <math>L'(x)=8-\frac{2500}{x^2}= \frac{8x^2-2500}{x^2}.</math>
|-
|The denominator can never be zero, and if we set the numerator to zero we find
|-
|     <math>x=\pm\sqrt{\frac{2500}{8}} = \pm\frac{50}{2\sqrt{2}} =\pm\frac{\,\,25}{\sqrt{2}}.</math>
|-
|Of course, we can't have negative fencing lengths, so we can ignore the negative root. Finally, we use the area relation to find
|-
|     <math>y =\frac{500}{25\sqrt{2}} = \frac{500\sqrt{2}}{25} = 20\sqrt{2}. </math>
|-
|Thus, the least amount of fencing is used when we size our <math style="vertical-align: 0%;">500</math> sq. ft. pens as <math style="vertical-align: -8%;">20\sqrt{2}</math> feet by <math style="vertical-align: -22%;">25/\sqrt{2}</math> feet.
|}

[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 50: / Line 50: @@
 |The pasture should be 20 meters by 10 meters.
 |}
-[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]
+[[022_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 27: / Line 27: @@
 !Step 3: &nbsp;
 |-
-|'''Optimize:''' &nbsp; Since <math style="vertical-align: -4px">xy=200</math>, we also know <math style="vertical-align: -4px;">y=200/x</math>. Plugging this into our equation for length, we have
+|'''Optimize:''' &nbsp; Since <math style="vertical-align: -4px">xy=200</math>, we also know <math style="vertical-align: -5px;">y=200/x</math>. Plugging this into our equation for length, we have
 |-
+|
@@ Line 42: / Line 42: @@
 ::<math>y\,=\,\frac{200}{x}\,=\,\frac{200}{10}\,=\, 20. </math>
 |-
-|Thus, the least amount of fencing is used when we size our <math style="vertical-align: 0%;">200</math> sq. m. pasture as <math style="vertical-align: 0px">20</math> meters by <math style="vertical-align: 0px">10</math> meters.
+|Thus, the least amount of fencing is used when we size our <math style="vertical-align: 0%;">200</math> sq. m. pasture as <math style="vertical-align: 0px">20</math> meters by <math style="vertical-align: -2px">10</math> meters.
 |}

@@ Line 13: / Line 13: @@
 !Step 1: &nbsp;
 |-
-|'''Declare Variables:''' &nbsp; We are attempting to enclose a rectangular area, such that we use as little fencing as possible for the three fenced sides.  Let's use <math style="vertical-align: 0%;">x</math> and <math style="vertical-align: -18%;">y</math> as indicated in the image, and simply call the length of fencing required <math style="vertical-align: 1%;">L</math>.
+|'''Declare Variables:''' &nbsp; We are attempting to enclose a rectangular area, such that we use as little fencing as possible for the three fenced sides.  Let's use <math style="vertical-align: 0px">x</math> and <math style="vertical-align: -4px;">y</math> as indicated in the image, and simply call the length of fencing required <math style="vertical-align: 0px">L</math>.
 |}
@@ Line 19: / Line 19: @@
 !Step 2: &nbsp;
 |-
-|'''Form the Equations:''' &nbsp; With the variables declared, we need two fences of length <math style="vertical-align: 0px">x</math> and one fence of length <math style="vertical-align: 0px;">y</math>.  Together, we need a total length of <math style="vertical-align: -3px%;">L=2x+y</math>.
+|'''Form the Equations:''' &nbsp; With the variables declared, we need two fences of length <math style="vertical-align: 0px">x</math> and one fence of length <math style="vertical-align: -4px;">y</math>.  Together, we need a total length of <math style="vertical-align: -4px">L=2x+y</math>.
 |-
-|On the other hand, we know that the pasture has a fixed area of <math style="vertical-align: 0%;">200</math> square meters.  Thus, we also know that <math style="vertical-align: -3px;">xy=200</math>.
+|On the other hand, we know that the pasture has a fixed area of <math style="vertical-align: 0px">200</math> square meters.  Thus, we also know that <math style="vertical-align: -4px;">xy=200</math>.
 |}
@@ Line 27: / Line 27: @@
 !Step 3: &nbsp;
 |-
-|'''Optimize:''' &nbsp; Since <math style="vertical-align: -18%;">xy=500</math>, we also know <math style="vertical-align: -21%;">y=500/x</math>. Plugging this into our equation for length, we have
+|'''Optimize:''' &nbsp; Since <math style="vertical-align: -4px">xy=200</math>, we also know <math style="vertical-align: -4px;">y=200/x</math>. Plugging this into our equation for length, we have
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>L(x) = 8x + 5\cdot \frac{500}{x} = 8x + \frac{2500}{x}.</math>
 |-
 |We now take the derivative to find
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>L'(x)=8-\frac{2500}{x^2}= \frac{8x^2-2500}{x^2}.</math>
 |-
-|The denominator can never be zero, and if we set the numerator to zero we find
+|This has roots <math style="vertical-align: 0px">\pm 10</math>.   Of course, we can't have negative fencing lengths, so we can ignore the negative root.  Finally, we use the area relation to find
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>x=\pm\sqrt{\frac{2500}{8}} = \pm\frac{50}{2\sqrt{2}} =\pm\frac{\,\,25}{\sqrt{2}}.</math>
 |-
-|Of course, we can't have negative fencing lengths, so we can ignore the negative root.  Finally, we use the area relation to find
+|Thus, the least amount of fencing is used when we size our <math style="vertical-align: 0%;">200</math> sq. m. pasture as <math style="vertical-align: 0px">20</math> meters by <math style="vertical-align: 0px">10</math> meters.
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>y =\frac{500}{25\sqrt{2}} = \frac{500\sqrt{2}}{25} = 20\sqrt{2}. </math>
+|The pasture should be 20 meters by 10 meters.
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]