022 Sample Final A, Problem 9 - Revision history

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MathAdmin: Created page with " Given demand $p = 116 - 3x$ , and cost $C = x^2 + 20x + 64$ , fi..."

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Created page with " Given demand <math style="vertical-align: -3.25px">p = 116 - 3x</math>, and cost <math style="vertical-align: -2.1px">C = x^2 + 20x + 64</math>, fi..."

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Given demand <math style="vertical-align: -3.25px">p = 116 - 3x</math>, and cost  <math style="vertical-align: -2.1px">C = x^2 + 20x + 64</math>, 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).

{| 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: -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; 
*<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; 
*<math style="vertical-align: -20%">\overline{C}(x)</math>, the '''average cost''' of producing <math style="vertical-align: 0%">x</math> units. 
|-
|In particular, we have the relations
|-
|
::<math>P(x)\,=\,R(x)-C(x),</math>
|-
|while
|-
|
::<math>R(x)\,=\,x\cdot p(x).</math>
|-
|and
|-
|
::<math>\overline{C}(x)\,=\,\frac{C(x)}{x}.</math>
|-
|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.

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:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a):  
|-
|The revenue function is
::<math>R(x)\,=\, x\cdot p(x)\,=\, x(116-3x)\,=\,116x-3x^{2}</math>.
Thus, the marginal revenue at a production level of <math style="vertical-align: 0px">7</math> units is simply
::<math>R'(7)\,=\,116-6x\bigg|_{x=7}\,=\,116-6(7)\,=\,74.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b):  
|-
|We have that the average cost function is
 
::<math>\begin{array}{rcl}
\overline{C}(x) & = & {\displaystyle {\displaystyle \frac{C(x)}{x}}}\\
\\
& = & {\displaystyle {\displaystyle \frac{x^{2}+20x+64}{x}}}\\
\\
& = & {\displaystyle x+20+\frac{64}{x}.}
\end{array}</math>

Our first derivative is then

::<math>\overline{C}\,'(x)\,=\,1-\frac{64}{x^{2}}\,=\,\frac{x^{2}-64}{x^{2}}\,=\,\frac{(x-8)(+8)}{x^{2}}.</math>

This has a single positive root at <math style="vertical-align: 0px">x=8</math>, which will correspond to the minimum average cost.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(c):  
|-
|First, we find the equation for profit. Using part of (a), we have

::<math>\begin{array}{rcl}
P(x) & = & {\displaystyle {\displaystyle R(x)-C(x)}}\\
\\
& = & 116x-3x^{2}-(x^{2}+20x+64)\\
\\
& = & -4x^{2}+136x+64.
\end{array}</math>

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

::<math>0\,=\,P'(x)\,=\,-8x+136\,=\,-8(x-17),</math>

which has a root at <math style="vertical-align: -1px">x=17</math>. Plugging this into our function for profit,
we have

::<math>P(17)\,=\,-4(17)^{2}+136(17)+64\,=\,1220.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)''' The marginal revenue at a production level of <math style="vertical-align: 0px">7</math> units is <math style="vertical-align: -1px">74</math>.
|-
|'''(b)''' The minimum average cost occurs at a production level of <math style="vertical-align: 0px">8</math> units.
|-
|'''(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.
|-
|Note that monetary units were not provided in the statement of the problem.
|}

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

@@ Line 78: / Line 78: @@
   & = & 116x-3x^{2}-(x^{2}+20x+64)\\
 \\
-  & = & -4x^{2}+96x+64.
+  & = & -4x^{2}+96x-64.
 \end{array}</math>
@@ Line 88: / Line 88: @@
 we have
-::<math>P(12)\,=\,-4(12)^{2}+96(12)+64\,=\,640.</math>
+::<math>P(12)\,=\,-4(12)^{2}+96(12)-64\,=\,512.</math>
 |}
@@ Line 101: / Line 101: @@
 |-
+|
-:'''(c)''' The maximum profit of <math style="vertical-align: -1px">640</math> occurs at a production level of <math style="vertical-align: -1px">12</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.

@@ Line 63: / Line 63: @@
 Our first derivative is then
-::<math>\overline{C}\,'(x)\,=\,1-\frac{64}{x^{2}}\,=\,\frac{x^{2}-64}{x^{2}}\,=\,\frac{(x-8)(+8)}{x^{2}}.</math>
+::<math>\overline{C}\,'(x)\,=\,1-\frac{64}{x^{2}}\,=\,\frac{x^{2}-64}{x^{2}}\,=\,\frac{(x-8)(x+8)}{x^{2}}.</math>
 This has a single positive root at <math style="vertical-align: 0px">x=8</math>, which will correspond to the minimum average cost.

@@ Line 52: / Line 52: @@
 |-
 |We have that the average cost function is
-<br>
 ::<math>\begin{array}{rcl}
 \overline{C}(x) & = & {\displaystyle {\displaystyle \frac{C(x)}{x}}}\\
@@ Line 94: / Line 94: @@
 !Final Answer: &nbsp;
 |-
 |'''(a)''' The marginal revenue at a production level of <math style="vertical-align: 0px">7</math> units is <math style="vertical-align: -1px">74</math>.
 |-
 |'''(b)''' The minimum average cost occurs at a production level of <math style="vertical-align: 0px">8</math> units.
 |-
 |'''(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.
 |-
 |Note that monetary units were not provided in the statement of the problem.

@@ Line 32: / Line 32: @@
 ::<math>\overline{C}(x)\,=\,\frac{C(x)}{x}.</math>
 |-
-|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.
+|The '''marginal profit''' at <math style="vertical-align: -3px">x_0</math> units is defined to be the effective profit of the next unit produced, and is precisely <math style="vertical-align: -5px">P'(x_0)</math>.  Similarly, the '''marginal revenue''' or '''marginal cost''' would be <math style="vertical-align: -5px">R'(x_0)</math> or &thinsp;<math style="vertical-align: -5px">C'(x_0)</math>, 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.

@@ Line 25: / Line 25: @@
 |-
+|
-::<math>R(x)\,=\,x\cdot p(x).</math>
+::<math>R(x)\,=\,x\cdot p(x),</math>
 |-
 |and