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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!

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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