Difference between revisions of "022 Sample Final A, Problem 9"

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  & = & 116x-3x^{2}-(x^{2}+20x+64)\\
 
  & = & 116x-3x^{2}-(x^{2}+20x+64)\\
 
\\
 
\\
  & = & -4x^{2}+136x+64.
+
  & = & -4x^{2}+96x-64.
 
\end{array}</math>
 
\end{array}</math>
  
 
To find the maximum value, we need to find a root of the derivative:
 
To find the maximum value, we need to find a root of the derivative:
  
::<math>0\,=\,P'(x)\,=\,-8x+136\,=\,-8(x-17),</math>
+
::<math>0\,=\,P'(x)\,=\,-8x+96\,=\,-8(x-12),</math>
  
which has a root at <math style="vertical-align: -1px">x=17</math>. Plugging this into our function for profit,
+
which has a root at <math style="vertical-align: -1px">x=12</math>. Plugging this into our function for profit,
 
we have
 
we have
  
::<math>P(17)\,=\,-4(17)^{2}+136(17)+64\,=\,1220.</math>
+
::<math>P(12)\,=\,-4(12)^{2}+96(12)-64\,=\,512.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
:'''(c)''' The maximum profit of <math style="vertical-align: -1px">1220</math> occurs at a production level of <math style="vertical-align: -1px">17</math> units.  
+
:'''(c)''' The maximum profit of <math style="vertical-align: -1px">512</math> &thinsp; occurs at a production level of <math style="vertical-align: -1px">12</math> units.  
 
|-
 
|-
 
|Note that monetary units were not provided in the statement of the problem.
 
|Note that monetary units were not provided in the statement of the problem.

Latest revision as of 18:13, 5 December 2016

Given demand , and cost  , find:

a) Marginal revenue when x = 7 units.
b) The quantity (x-value) that produces minimum average cost.
c) Maximum profit (find both the x-value and the profit itself).
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;
  • , the average cost of producing units.
In particular, we have the relations
while
and
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.

On the other hand, any time they speak of minimizing or maximizing, we need to find a local extrema. These occur when the first derivative is zero.

 Solution:

(a):  
The revenue function is
.

Thus, the marginal revenue at a production level of units is simply

(b):  
We have that the average cost function is

Our first derivative is then

This has a single positive root at , which will correspond to the minimum average cost.

(c):  
First, we find the equation for profit. Using part of (a), we have

To find the maximum value, we need to find a root of the derivative:

which has a root at . Plugging this into our function for profit, we have

Final Answer:  
(a) The marginal revenue at a production level of units is .
(b) The minimum average cost occurs at a production level of units.
(c) The maximum profit of   occurs at a production level of units.
Note that monetary units were not provided in the statement of the problem.


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