Science:Math Exam Resources/Courses/MATH104/December 2012/Question 04 (a)

MATH104 December 2012

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Question 04 (a)
The demand equation for a certain product is $\displaystyle {}q^{2}+p^{3/2}+2p=20$ where q is the number of units per hour the manufacturer can sell at a price of p dollars per unit. Calculate q and ${\frac {{\textrm {d}}q}{{\textrm {d}}p}}$ when the price is set at $4.

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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
To calculate $dq/dp$ you can use implicit differentiation. Alternatively, you can solve for $q$ in terms of $p$ and then differentiate.

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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. To determine $q$ when $p=4$ is simply a matter of substituting $p=4$ into the demand equation and solving for $q$. ${\begin{aligned}q^{2}+p^{3/2}+2p&=20\\q^{2}+8+8&=20\\q^{2}&=4\\q&=\pm 2.\end{aligned}}$ We ignore the negative solution since it is not possible to have negative demand. Therefore ${\color {blue}q=2}.$ To determine $dq/dp$, we use implicit differentiation. ${\begin{aligned}{\frac {d}{dp}}\left(q^{2}+p^{3/2}+2p\right)&={\frac {d}{dp}}\left(20\right)\\2q{\frac {dq}{dp}}+{\frac {3}{2}}p^{1/2}+2&=0\\{\frac {dq}{dp}}&={\frac {1}{2q}}\left(-2-{\frac {3}{2}}p^{1/2}\right)\end{aligned}}$ Substituting the values $p=4,\,q=2$ into the expression gives the value of $dq/dp$: ${\begin{aligned}{\frac {dq}{dp}}&={\frac {1}{4}}\left(-2-3\right)\quad \rightarrow \quad {\color {blue}{\frac {dq}{dp}}}\,&{\color {blue}=-{\frac {5}{4}}}\end{aligned}}$

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. Instead of implicitly differentiating, we first solve the equation for q and then differentiate normally. ${\begin{aligned}q^{2}+p^{3/2}+2p&=20\\q^{2}&=20-p^{3/2}-2p\\q&={\sqrt {20-p^{3/2}-2p}}\end{aligned}}$ Then ${\frac {dq}{dp}}={\frac {-(3/2)p^{1/2}-2}{2{\sqrt {20-p^{3/2}-2p}}}}$ Substituting in $p=4$ gives ${\frac {dq}{dp}}={\frac {-(3/2)(4)^{1/2}-2}{2{\sqrt {20-(4)^{3/2}-2(4)}}}}={\frac {-(3/2)(2)-2}{2{\sqrt {20-8-8}}}}={\frac {-5}{4}}$

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