Math 22 Business and Economics Applications

Optimization in Business and Economics

1) Find the number of units $x$ that minimizes the average cost per unit ${\overline {C}}$ when $C=2x^{2}+348x+7200$

Solution:
Notice: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\overline {C}}={\frac {C}{x}}={\frac {2x^{2}+348x+7200}{x}}=2x+348+{\frac {7200}{x}}}
Then, ${\overline {C}}'=2-{\frac {7200}{x^{2}}}=0$, so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x^{2}=3600} , so $x=\pm {\sqrt {3600}}=\pm 60=60$ since $x$ is positive.

2) Find the price that will maximize profit for the demand and cost functions, where $p$ is the price, $x$ is the number of units, and $C$ is the cost. Given the demand function Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle p(x)=90-x} and the cost function $C(x)=100+30x$.

Solution:
Notice: The revenue function $R(x)=x\cdot p(x)=x(90-x)=90x-x^{2}$
The Profit function is $P(x)=R(x)-C(x)=90x-x^{2}-(100+30x)=90x-x^{2}-100-30x=-x^{2}+60x-100$
Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle P'(x)=-2x+60=0} , so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=30}
So, $p(30)=90-30=60$
Therefore, the price is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \$60} a unit will maximize the profit.
Then, ${\overline {C}}'=2-{\frac {7200}{x^{2}}}=0$, so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x^{2}=3600} , so $x=\pm {\sqrt {3600}}=\pm 60=60$ since $x$ is positive.

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