Difference between revisions of "Math 22 Business and Economics Applications"
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|Notice: <math>\overline{C}=\frac{C}{x}=\frac{2x^2+348x+7200}{x}=2x+348+\frac{7200}{x}</math> | |Notice: <math>\overline{C}=\frac{C}{x}=\frac{2x^2+348x+7200}{x}=2x+348+\frac{7200}{x}</math> | ||
+ | |- | ||
+ | |Then, <math>\overline{C} '=2-\frac{7200}{x^2}=0</math>, so <math>x^2=3600</math>, so <math>x=\pm\sqrt{3600}=\pm 60=60</math> since <math>x</math> is positive. | ||
+ | |} | ||
+ | |||
+ | '''2)''' Find the price that will maximize profit for the demand and cost functions, where <math>p</math> is the price, <math>x</math> is the number of units, and <math>C</math> is the cost. Given the demand function <math>p(x)=90-x</math> and the cost function <math>C(x)=100+30x</math>. | ||
+ | |||
+ | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | !Solution: | ||
+ | |- | ||
+ | |Notice: The revenue function <math>R(x)=x\cdot p(x)=x(90-x)=90x-x^2</math> | ||
+ | |- | ||
+ | |The Profit function is <math>P(x)=R(x)-C(x)=90x-x^2-(100+30x)=90x-x^2-100-30x=-x^2+60x-100</math> | ||
+ | |- | ||
+ | |Then, <math>P'(x)=-2x+60=0</math>, so <math>x=30</math> | ||
+ | |- | ||
+ | |So, <math>p(30)=90-30=60</math> | ||
+ | |- | ||
+ | |Therefore, the price is <math>\$ 60</math> a unit will maximize the profit. | ||
|- | |- | ||
|Then, <math>\overline{C} '=2-\frac{7200}{x^2}=0</math>, so <math>x^2=3600</math>, so <math>x=\pm\sqrt{3600}=\pm 60=60</math> since <math>x</math> is positive. | |Then, <math>\overline{C} '=2-\frac{7200}{x^2}=0</math>, so <math>x^2=3600</math>, so <math>x=\pm\sqrt{3600}=\pm 60=60</math> since <math>x</math> is positive. |
Revision as of 06:51, 2 August 2020
Optimization in Business and Economics
1) Find the number of units that minimizes the average cost per unit when
Solution: |
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Notice: |
Then, , so , so since is positive. |
2) Find the price that will maximize profit for the demand and cost functions, where is the price, is the number of units, and is the cost. Given the demand function and the cost function .
Solution: |
---|
Notice: The revenue function |
The Profit function is |
Then, , so |
So, |
Therefore, the price is a unit will maximize the profit. |
Then, , so , so since is positive. |
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