Difference between revisions of "022 Exam 1 Sample A, Problem 5"

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Find the marginal revenue and marginal profit at <math style="vertical-align: -3%">x=4</math>, given the demand function  
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<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>  
 
::<math>p=\frac{200}{\sqrt{x}}</math>  
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! Foundations: &nbsp;  
 
! 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>.
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|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:
 
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>
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*<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>
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*<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>
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*<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
 
|In particular, we have the relations
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::<math>R(x)=x\cdot p(x).</math>
 
::<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.
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|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.
  
 
|}
 
|}
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!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
|-
 
|-
|'''Write the Basic Equation:'''  
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|'''Find the Important Functions:''' We have
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|-
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|
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::<math>R(x)\,\,=\,\,x\cdot p(x)\,\,=\,\,x\cdot \frac{200}{\sqrt {x}}\,\,=\,\,200 \sqrt{x}.</math>
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|-
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|From this,
 +
|-
 +
|
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::<math>P(x)\,\,=\,\,R(x)-C(x)\,\,=\,\,200 \sqrt{x}- \left( 100+15x+3x^{2} \right) .</math>
 
|}
 
|}
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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|-
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|'''Find the Marginal Revenue and Profit:''' The equation for marginal revenue is
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|-
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|
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::<math>R'(x)\,\,=\,\,\left(200 \sqrt{x}\right) '\,\,=\,\,200\cdot \frac{1}{2}\cdot\frac{1}{\sqrt{x}}\,\,=\,\,\frac{100}{\sqrt{x}}, </math>
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|-
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|while the equation for marginal profit is
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|-
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|
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::<math>P'(x)\,\,=\,\,R'(x)-C'(x)\,\,=\,\,\frac{100}{\sqrt{x}}-(15+6x).</math>
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|-
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|At <math style="vertical-align: -3%">x=4</math>, we find the marginal revenue is
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|-
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|
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::<math>R'(4)\,\,=\,\,\frac{100}{\sqrt{4}}\,\,=\,\,50. </math>
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|-
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|On the other hand, marginal profit is
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|-
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|
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::<math>P'(4)\,\,=\,\,\frac{100}{\sqrt{4}}-(15+6(4))\,\,=\,\,50-39\,\,=\,\,11.</math>
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|-
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|Thus, it is profitable to produce another item.
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;
 
!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.
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|
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::<math>R'(4)\,\,=\,\,50;\qquad P'(4)\,\,=\,\,11. </math>
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|-
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|Thus, it is profitable to produce another item.
 
|}
 
|}
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[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 11:39, 20 April 2015

Find the marginal revenue and marginal profit at , given the demand function

and the cost function

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

Foundations:  
Recall that the demand function, , relates the price per unit to the number of units sold, .

Moreover, we have several important important functions:

  • , the total cost to produce units;
  • , the total revenue (or gross receipts) from producing units;
  • , the total profit from producing units.
In particular, we have the relations
and
Finally, the marginal profit at units is defined to be the effective profit of the next unit produced, and is precisely . Similarly, the marginal revenue or marginal cost would be or , respectively.

 Solution:

Step 1:  
Find the Important Functions: We have
From this,
Step 2:  
Find the Marginal Revenue and Profit: The equation for marginal revenue is
while the equation for marginal profit is
At , we find the marginal revenue is
On the other hand, marginal profit is
Thus, it is profitable to produce another item.
Final Answer:  
Thus, it is profitable to produce another item.


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