Science:Math Exam Resources/Courses/MATH104/December 2016/Question 10

MATH104 December 2016

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Question 10
A company’s demand and cost curves are given by: $p+{\sqrt {q}}=150\quad {\text{ and }}\quad C(q)=2500+6q.$ Determine the selling price which would produce the most profit. To receive full credit you must justify your solution and simplify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
First find the function $P=P(q)$ for the profit. Then, find the quantity $q$ at which $P$ has the global maximum.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First we find the function $P=P(q)$ for the profit. Using ${\text{Profit}}={\text{Revenue}}-{\text{Cost}}\quad {\text{and}}\quad {\text{Revenue}}={\text{Price}}\times {\text{Quantity}}=pq$, we obtain $P(q)=pq-C(q)=q(150-{\sqrt {q}})-2500-6q=150q-q^{\frac {3}{2}}-2500-6q$. To make the most profit, we find the derivative of $P$ $P'=150-{\frac {3}{2}}q^{\frac {1}{2}}-6$. Then, set $P'=0$ to get $150-{\frac {3}{2}}q^{\frac {1}{2}}-6=0,\implies 144={\frac {3}{2}}q^{\frac {1}{2}}\implies {\sqrt {q}}=96$. To see $P$ has global maximum at the critical point $q=96^{2}$, we consider its second derivative; $P''=-{\frac {3}{4}}q^{-{\frac {1}{2}}}<0$. This follows that $P(q)$ is always concave down, making its critical value as the global maximum. Finally, to find the selling price which produces the most profit, we substitute ${\sqrt {q}}=96$ into the the demand curve; $p+96=150\implies p=54$. Answer: $\color {blue}p=54$