009A Sample Final A, Problem 7 - Revision history

MathAdmin at 16:53, 2 April 2015

2015-04-02T16:53:46Z

MathAdmin at 03:48, 28 March 2015

2015-03-28T03:48:45Z

MathAdmin at 03:48, 28 March 2015

2015-03-28T03:48:11Z

MathAdmin at 03:47, 28 March 2015

2015-03-28T03:47:06Z

MathAdmin at 03:43, 28 March 2015

2015-03-28T03:43:50Z

MathAdmin at 21:09, 26 March 2015

2015-03-26T21:09:28Z

MathAdmin at 16:53, 26 March 2015

2015-03-26T16:53:38Z

MathAdmin at 20:03, 24 March 2015

2015-03-24T20:03:25Z

MathAdmin at 20:03, 24 March 2015

2015-03-24T20:03:00Z

MathAdmin at 19:56, 24 March 2015

2015-03-24T19:56:34Z

@@ Line 21: / Line 21: @@
 !Step 2: &nbsp;
 |-
-|'''Form the Equations:''' &nbsp; Notice that we need fencing between each of the pens (think "lion-antelope-lion-antelope" if this isn't clear). We require <math style="vertical-align: 0%;">2</math> pieces of length <math style="vertical-align: 0%;">x</math> for each pen, and a total of <math style="vertical-align: 0%;">5</math> pieces of length <math style="vertical-align: -18%;">y</math>. Together, we need a total length of <math style="vertical-align: -18%;">L=8x+5y</math>.
+|'''Form the Equations:''' &nbsp; Notice that we need fencing between each of the pens, so we require a total of <math style="vertical-align: 0%;">5</math> pieces of length <math style="vertical-align: -18%;">y</math>. We also need <math style="vertical-align: 0%;">2</math> pieces of length <math style="vertical-align: 0%;">x</math> for each pen, which means there are a total of <math style="vertical-align: -5%;">4\cdot 2</math> &thinsp; pieces required of length <math style="vertical-align: 0%;">x</math>. Together, we need a total length of <math style="vertical-align: -18%;">L=8x+5y</math>.
 |-
 |On the other hand, we know that each pen has a fixed area of <math style="vertical-align: 0%;">500</math> square feet.  Thus, we also know that <math style="vertical-align: -18%;">xy=500</math>.

@@ Line 29: / Line 29: @@
 !Step 3: &nbsp;
 |-
-|'''Optimize:''' &nbsp; Since <math style="vertical-align: -18%;">xy=500</math>, we also know <math style="vertical-align: -22%;">y=500/x</math>. Plugging this into our equation for length, we have
+|'''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
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>L(x) = 8x + 5\cdot \frac{500}{x} = 8x + \frac{2500}{x}.</math>
@@ Line 45: / Line 45: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <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: -23%;">25/\sqrt{2}</math> feet.
+|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 29: / Line 29: @@
 !Step 3: &nbsp;
 |-
-|'''Optimize:''' &nbsp; Since <math style="vertical-align: -18%;">xy=500</math>, we also know <math style="vertical-align: -23%;">y=500/x</math>. Plugging this into our equation for length, we have
+|'''Optimize:''' &nbsp; Since <math style="vertical-align: -18%;">xy=500</math>, we also know <math style="vertical-align: -22%;">y=500/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>
@@ Line 45: / Line 45: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <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: -24%;">25/\sqrt{2}</math> feet.
+|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: -23%;">25/\sqrt{2}</math> feet.
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 15: / Line 15: @@
 !Step 1: &nbsp;
 |-
-|'''Declare Variables:''' &nbsp; We are attempting to find the dimensions of a single pen, such that we use as little fencing as possible for all four pens.  Let's use <math style="vertical-align: 0%;">x</math> and <math style="vertical-align: -20%;">y</math> as indicated in the image, and simply call the length of fencing required <math style="vertical-align: 0%;">L</math>.
+|'''Declare Variables:''' &nbsp; We are attempting to find the dimensions of a single pen, such that we use as little fencing as possible for all four pens.  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>.
 |}
@@ Line 21: / Line 21: @@
 !Step 2: &nbsp;
 |-
-|'''Form the Equations:''' &nbsp; Notice that we need fencing between each of the pens (think "lion-antelope-lion-antelope" if this isn't clear). We require <math style="vertical-align: 0%;">2</math> pieces of length <math style="vertical-align: 0%;">x</math> for each pen, and a total of <math style="vertical-align: 0%;">5</math> pieces of length <math style="vertical-align: -20%;">y</math>. Together, we need a total length of <math style="vertical-align: -20%;">L=8x+5y</math>.
+|'''Form the Equations:''' &nbsp; Notice that we need fencing between each of the pens (think "lion-antelope-lion-antelope" if this isn't clear). We require <math style="vertical-align: 0%;">2</math> pieces of length <math style="vertical-align: 0%;">x</math> for each pen, and a total of <math style="vertical-align: 0%;">5</math> pieces of length <math style="vertical-align: -18%;">y</math>. Together, we need a total length of <math style="vertical-align: -18%;">L=8x+5y</math>.
 |-
-|On the other hand, we know that each pen has a fixed area of <math style="vertical-align: 0%;">500</math> square feet.  Thus, we also know that <math style="vertical-align: -20%;">xy=500</math>.
+|On the other hand, we know that each pen has a fixed area of <math style="vertical-align: 0%;">500</math> square feet.  Thus, we also know that <math style="vertical-align: -18%;">xy=500</math>.
 |}
@@ Line 29: / Line 29: @@
 !Step 3: &nbsp;
 |-
-|'''Optimize:''' &nbsp; Since <math style="vertical-align: -20%;">xy=500</math>, we also know <math style="vertical-align: -25%;">y=500/x</math>. Plugging this into our equation for length, we have
+|'''Optimize:''' &nbsp; Since <math style="vertical-align: -18%;">xy=500</math>, we also know <math style="vertical-align: -23%;">y=500/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>
@@ Line 45: / Line 45: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <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: -10%;">20\sqrt{2}</math> feet by <math style="vertical-align: -26%;">25/\sqrt{2}</math> feet.
+|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: -24%;">25/\sqrt{2}</math> feet.
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 15: / Line 15: @@
 !Step 1: &nbsp;
 |-
-|'''Declare Variables:''' &nbsp; We are attempting to find the dimensions of a single pen, such that we use as little fencing as possible for all four pens.  Let's use ''x'' and ''y'' as indicated in the image, and simply call the length of fencing required ''L''.
+|'''Declare Variables:''' &nbsp; We are attempting to find the dimensions of a single pen, such that we use as little fencing as possible for all four pens.  Let's use <math style="vertical-align: 0%;">x</math> and <math style="vertical-align: -20%;">y</math> as indicated in the image, and simply call the length of fencing required <math style="vertical-align: 0%;">L</math>.
 |}
@@ Line 21: / Line 21: @@
 !Step 2: &nbsp;
 |-
-|'''Form the Equations:''' &nbsp; Notice that we need fencing between each of the pens (think "lion-antelope-lion-antelope" if this isn't clear). We require 2 pieces of length ''x'' for each pen, and a total of 5 pieces of length ''y''. Together, we need a total length of ''L'' = 8''x'' + 5''y''.
+|'''Form the Equations:''' &nbsp; Notice that we need fencing between each of the pens (think "lion-antelope-lion-antelope" if this isn't clear). We require <math style="vertical-align: 0%;">2</math> pieces of length <math style="vertical-align: 0%;">x</math> for each pen, and a total of <math style="vertical-align: 0%;">5</math> pieces of length <math style="vertical-align: -20%;">y</math>. Together, we need a total length of <math style="vertical-align: -20%;">L=8x+5y</math>.
 |-
-|On the other hand, we know that each pen has a fixed area of 500 square feet.  Thus, we also know that ''xy'' = 500.
+|On the other hand, we know that each pen has a fixed area of <math style="vertical-align: 0%;">500</math> square feet.  Thus, we also know that <math style="vertical-align: -20%;">xy=500</math>.
 |}
@@ Line 29: / Line 29: @@
 !Step 3: &nbsp;
 |-
-|'''Optimize:''' &nbsp; Since ''xy'' = 500, we also know ''y'' = 500/''x''. Plugging this into our equation for length, we have
+|'''Optimize:''' &nbsp; Since <math style="vertical-align: -20%;">xy=500</math>, we also know <math style="vertical-align: -25%;">y=500/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>
@@ Line 45: / Line 45: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <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 500 sq. ft. pens as 20&radic;<span style="text-decoration:overline">2</span> feet by 25/&radic;<span style="text-decoration:overline">2</span> feet.
+|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: -10%;">20\sqrt{2}</math> feet by <math style="vertical-align: -26%;">25/\sqrt{2}</math> feet.
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 |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:'''
+&nbsp;'''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
 <br>
 <span style="font-size:135%"><font face=Times Roman>  7. A farmer wishes to make 4 identical rectangular pens, each with
@@ Line 12: / Line 13: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Part (a): &nbsp;
+!Step 1: &nbsp;
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]