Q.137. Application of derivatives

Question 137.
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 + 7p^2 = 1000
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.

Answer 137.
x^2 + 2xp + 7p^2 = 1000
Differentiating with respect to time,
2x dx/dt + 2 p dx/dt + 2x dp/dt + 14 p dp/dt = 0

Rate of price increase, dp/dt = $ 0.20 per month when p = $ 10.
Put p = 10 in the given equation to find x.
x^2 + 20x + 700 = 1000
=> x^2 + 20 x - 300 = 0
=> (x + 30)(x - 10) = 0
=> x = 10 ( x = - 30 is not possible)

Using this value of x, p and dp/dt in the differentiated eqn.,
2(10) dx/dt + 2(10) dx/dt + 2(10)(0.2) + 14(10)(0.2) = 0
=> 40 dx/dt = - 32
=> dx/dt = - 32/40 * (100) = - 80
means decrease in rate of demand by 80 per month.
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