Science:Math Exam Resources/Courses/MATH104/December 2016/Question 06 (b)

MATH104 December 2016

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Question 06 (b)
A company's demand curve is given by $p+10q=168$ . Determine the selling price which would produce the most revenue.

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Hint
Revenue equals the price times the number of products that are sold.

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Solution 1
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From demand curve, we have the relationship between $p$ and $q$ as $q={\frac {168-p}{10}}$ . Following the hint, the revenue is $R=p\cdot q$ and hence ${\begin{aligned}R&=p\cdot {\frac {168-p}{10}}\\\end{aligned}}$ If we want to find the price that makes most revenue, then that point should be at the maximum of function $R$ , where the derivative of function is zero. Therefore setting ${\frac {dR}{dp}}=1\cdot {\frac {168-p}{10}}+p\cdot {\frac {-1}{10}}={\frac {1}{10}}\cdot (168-p-p)=0$ gives us $2p=168\rightarrow p=84$ Since ${\frac {d^{2}R}{dp^{2}}}=-{\frac {2}{10}}<0$ , the critical point $p=84$ is indeed the price which produces the most revenue. Answer: $\color {blue}84$

Solution 2
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Alt Sol: using demand elasticity formula $E={\frac {p}{q}}{\frac {dq}{dp}}$ $p+10q=168$ Consider $q$ is a function of $p$ , and we take the derivative of $p$ : $1+10{\frac {dq}{dp}}=0$ ${\frac {dq}{dp}}=-{\frac {1}{10}}$ Therefore $E=-{\frac {1}{10}}\cdot {\frac {p}{q}}$ Recall when $E=-1$ , we say the good is price unit elastic. This is the optimal price which means it maximizes revenue. So setting $E=-1$ gives us : $-1=-{\frac {1}{10}}\cdot {\frac {p}{q}}\rightarrow p=10q\rightarrow p=168-p\rightarrow p=84$ answer: $\color {blue}84$