Science:Math Exam Resources/Courses/MATH104/December 2014/Question 03 (c)

MATH104 December 2014

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Question 03 (c)
The price $p$ (in dollars) and the demand $q$ for a product are related by the following demand equation: $p^{3}+q+q^{3}=38$ Suppose the price increases at a rate of $\$7$/month. How fast does the demand decrease when the demand is $q=3$?

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What variable are you taking the derivative of demand with respect to (think about what unit of measurement is “months”)?

Hint 2
Implicitly differentiate with respect to time.

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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. From the previous part, we know that when $q=3$, by the relation give, $p=2$. Now, we take that above relation and differentiate it with respect to time: $3p^{2}{\frac {dp}{dt}}+{\frac {dq}{dt}}+3q^{2}{\frac {dq}{dt}}=0$ We know that ${\frac {dp}{dt}}=7,p=2,q=3$. Substitute into the new found relation and isolate for ${\frac {dq}{dt}}$ gives: ${\frac {dq}{dt}}={\frac {-3(2)^{2}(7)}{1+3(3)^{2}}}={\frac {-84}{28}}=-3$. So the demand is decreasing at a rate of $3$ units/month.

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