022 Exam 1 Sample A, Problem 5 - Revision history

MathAdmin at 19:39, 20 April 2015

2015-04-20T19:39:27Z

MathAdmin at 03:37, 14 April 2015

2015-04-14T03:37:34Z

MathAdmin at 03:13, 14 April 2015

2015-04-14T03:13:41Z

MathAdmin at 03:12, 14 April 2015

2015-04-14T03:12:02Z

MathAdmin: Created page with "Find the marginal revenue and marginal profit at $x=4$ , given the demand function :: $p=\frac{200}{\sqrt{x}}$
2015-04-14T02:56:20Z

Created page with "Find the marginal revenue and marginal profit at <math style="vertical-align: -3%">x=4</math>, given the demand function ::<math>p=\frac{200}{\sqrt{x}}</math> x=4</math>, given the demand function

::<math>p=\frac{200}{\sqrt{x}}</math>

and the cost function

::<math>C(x)=100+15x+3x^{2}.</math>

Should the firm produce one more item under these conditions? Justify
your answer.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|Recall that the demand function, <math style="vertical-align: -25%">p(x)</math>, relates the price per unit <math style="vertical-align: -20%">p</math> to the number of units sold, <math style="vertical-align: 0%">x</math>.
Moreover, we have several important important functions:
|-
|
<math style="vertical-align: -20%">C(x)</math>, the total cost to produce <math style="vertical-align: 0%">x</math> units; 
<math style="vertical-align: -20%">R(x)</math>, the total revenue (or gross receipts) from producing <math style="vertical-align: 0%">x</math> units; 
*<math style="vertical-align: -20%">P(x)</math>, the total profit from producing <math style="vertical-align: 0%">x</math> units. 
|-
|In particular, we have the relations
|-
|
::<math>P(x)=R(x)-C(x),</math>
|-
|and
|-
|
::<math>R(x)=x\cdot p(x).</math>
|-
|Finally, marginal profit at <math style="vertical-align: -20%">x_0</math> units is defined to be the effective cost of the next unit produced, and is precisely <math style="vertical-align: -25%">P'(x_0)</math>. Similarly, marginal revenue or cost would be <math style="vertical-align: -25%">R'(x_0)</math> or <math style="vertical-align: -25%">C'(x_0)</math>, respectively.

|}
 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|'''Write the Basic Equation:'''
|}

!Final Answer:  
|-
|With units, we have that the ladder is sliding down the wall at <math style="vertical-align: -25%">-3/2</math>  feet per second.
|}

[[022_Exam_1_Sample_A|'''Return to Sample Exam''']]

@@ Line 61: / Line 61: @@
 |-
+|
-::<math>P'(x)\,\,=\,\,R'(x)-C'(x)\,\,=\,\,\frac{100}{\sqrt{x}}-(30+6x).</math>
+::<math>P'(x)\,\,=\,\,R'(x)-C'(x)\,\,=\,\,\frac{100}{\sqrt{x}}-(15+6x).</math>
 |-
-|At <math style="vertical-align: -3%">x=4</math>, we find
+|At <math style="vertical-align: -3%">x=4</math>, we find the marginal revenue is
 |-
+|
@@ Line 71: / Line 71: @@
 |-
+|
-::<math>P'(4)\,\,=\,\,\frac{100}{\sqrt{4}}-(30+6x(4))\,\,=\,\,50-54\,\,=\,\,-4.</math>
+::<math>P'(4)\,\,=\,\,\frac{100}{\sqrt{4}}-(15+6(4))\,\,=\,\,50-39\,\,=\,\,11.</math>
 |-
-|Thus, it is <u>''not''</u> profitable to produce another item.
+|Thus, it is profitable to produce another item.
 |}
@@ Line 80: / Line 80: @@
 |-
+|
-::<math>R'(4)\,\,=\,\,50;\qquad P'(4)\,\,=\,\,-4. </math>
+::<math>R'(4)\,\,=\,\,50;\qquad P'(4)\,\,=\,\,11. </math>
 |-
-|Thus, it is <u>''not''</u> profitable to produce another item.
+|Thus, it is profitable to produce another item.
 |}
 [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 13: / Line 13: @@
 ! Foundations: &nbsp;
 |-
-|Recall that the demand function, <math style="vertical-align: -25%">p(x)</math>, relates the price per unit <math style="vertical-align: -17%">p</math> to the number of units sold, <math style="vertical-align: 0%">x</math>.
+|Recall that the '''demand function''', <math style="vertical-align: -25%">p(x)</math>, relates the price per unit <math style="vertical-align: -17%">p</math> to the number of units sold, <math style="vertical-align: 0%">x</math>.
 Moreover, we have several important important functions:
 |-
+|
-*<math style="vertical-align: -20%">C(x)</math>, the total cost to produce <math style="vertical-align: 0%">x</math> units;<br>
+*<math style="vertical-align: -20%">C(x)</math>, the '''total cost''' to produce <math style="vertical-align: 0%">x</math> units;<br>
-*<math style="vertical-align: -20%">R(x)</math>, the total revenue (or gross receipts) from producing <math style="vertical-align: 0%">x</math> units;<br>
+*<math style="vertical-align: -20%">R(x)</math>, the '''total revenue''' (or gross receipts) from producing <math style="vertical-align: 0%">x</math> units;<br>
-*<math style="vertical-align: -20%">P(x)</math>, the total profit from producing <math style="vertical-align: 0%">x</math> units.<br>
+*<math style="vertical-align: -20%">P(x)</math>, the '''total profit''' from producing <math style="vertical-align: 0%">x</math> units.<br>
 |-
 |In particular, we have the relations
@@ Line 31: / Line 31: @@
 ::<math>R(x)=x\cdot p(x).</math>
 |-
-|Finally, marginal profit at <math style="vertical-align: -20%">x_0</math> units is defined to be the effective cost of the next unit produced, and is precisely <math style="vertical-align: -22%">P'(x_0)</math>.  Similarly, marginal revenue or cost would be <math style="vertical-align: -22%">R'(x_0)</math> or <math style="vertical-align: -22%">C'(x_0)</math>, respectively.
+|Finally, the '''marginal profit''' at <math style="vertical-align: -20%">x_0</math> units is defined to be the effective profit of the next unit produced, and is precisely <math style="vertical-align: -22%">P'(x_0)</math>.  Similarly, the '''marginal revenue''' or '''marginal cost''' would be <math style="vertical-align: -22%">R'(x_0)</math> or <math style="vertical-align: -22%">C'(x_0)</math>, respectively.
 |}

@@ Line 51: / Line 51: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!Step 2: &nbsp;
 |-
-|'''Find the Marginal Revenue and Profit:''' The marginal revenue is
+|'''Find the Marginal Revenue and Profit:''' The equation for marginal revenue is
 |-
+|
 ::<math>R'(x)\,\,=\,\,\left(200 \sqrt{x}\right) '\,\,=\,\,200\cdot \frac{1}{2}\cdot\frac{1}{\sqrt{x}}\,\,=\,\,\frac{100}{\sqrt{x}}, </math>
 |-
-|while the marginal profit is
+|while the equation for marginal profit is
 |-
+|

@@ Line 1: / Line 1: @@
-Find the marginal revenue and marginal profit at <math style="vertical-align: -3%">x=4</math>, given the demand function
+<span class="exam">Find the marginal revenue and marginal profit at <math style="vertical-align: -3%">x=4</math>, given the demand function
 ::<math>p=\frac{200}{\sqrt{x}}</math>
@@ Line 13: / Line 13: @@
 ! Foundations: &nbsp;
 |-
-|Recall that the demand function, <math style="vertical-align: -25%">p(x)</math>, relates the price per unit <math style="vertical-align: -20%">p</math> to the number of units sold, <math style="vertical-align: 0%">x</math>.
+|Recall that the demand function, <math style="vertical-align: -25%">p(x)</math>, relates the price per unit <math style="vertical-align: -17%">p</math> to the number of units sold, <math style="vertical-align: 0%">x</math>.
 Moreover, we have several important important functions:
 |-
@@ Line 31: / Line 31: @@
 ::<math>R(x)=x\cdot p(x).</math>
 |-
-|Finally, marginal profit at <math style="vertical-align: -20%">x_0</math> units is defined to be the effective cost of the next unit produced, and is precisely <math style="vertical-align: -25%">P'(x_0)</math>.  Similarly, marginal revenue or cost would be <math style="vertical-align: -25%">R'(x_0)</math> or <math style="vertical-align: -25%">C'(x_0)</math>, respectively.
+|Finally, marginal profit at <math style="vertical-align: -20%">x_0</math> units is defined to be the effective cost of the next unit produced, and is precisely <math style="vertical-align: -22%">P'(x_0)</math>.  Similarly, marginal revenue or cost would be <math style="vertical-align: -22%">R'(x_0)</math> or <math style="vertical-align: -22%">C'(x_0)</math>, respectively.
 |}
@@ Line 39: / Line 39: @@
 !Step 1: &nbsp;
 |-
-|'''Write the Basic Equation:'''
+|'''Find the Important Functions:''' We have
 |}
 !Final Answer: &nbsp;
 |-
-|With units, we have that the ladder is sliding down the wall at <math style="vertical-align: -25%">-3/2</math>&thinsp; feet per second.
 |}
 [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]