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Calculus/Related Rates-Price and Demand

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Question
At a price of $p the demand x per month (in multiples of 100) for a new piece of software is given by
x^2+2xp+3p^2=600. Because of its popularity, the manufacturer is increasing the price at a rate of 20 cents per month. Find the corresponding rate of decrease in demand for the software when the software costs $10.

I am really close to finding the answer to this question. I feel like I'm just leaving one tiny part out because I can't seem to find x. I plugged in p to the original problem, but I'm unsure how to not get two answers for x. Thank you!

Answer
Let's run through the steps to make sure you have done it correctly.

dp/dt = 0.2

Differentiating x^2+2xp+3p^2=600 wrt x gives

2x + 2p + 2x (dp/dx) + 6p (dp/dx)= 0

(2x+6p)*(dp/dx) =  -2x-2p

dp/dx = (-x -p)/(x+3p)----------(1)

The chain rules gives us dp/dt = dp/dx * dx/dt   -------------(2)

When p=10, substituting into x^2+2xp+3p^2=600
gives  x^2 + 20x + 300 = 600

x^2 + 20x + 300 = 600

x^2 + 20x - 300 = 0

(x+10)^2 -300-10^2 =0

(x+10)^2 = 400

As such, x= sqrt(400) - 10 = 10      

or  x= -sqrt(400) -10 = -30 (which is rejected because obviously demand cannot be negative)

Substituting x=10 and p=10 into (1),

dp/dx = -20/40 = -1/2

Further substituting dp/dx= -1/2  and dp/dt = 0.2 into (2),

you should arrive at dx/dt = - 0.4  (in multiples of 100)

The corresponding rate of decrease in demand is therefore 0.4*100 = 40 pieces per month.

Bear in mind the demand (no. of pieces wanted) cannot be negative, whilst the rate of change of demand can be negative (which will mean a drop in sales).

Hope this helps. Peace.

Calculus

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