Question

A price p

 (in dollars) and demand x

 for a product are related by

2x^2-8xp+50p^2=7400

If the price is increasing at a rate of 2 dollars per month when the price is 10 dollars, find the rate of change of the demand.

Answer 1

$\displaystyle 2x^2 -8xp+2x^2=7400, \frac{dp}{dt} = 2\\ \text{Differentiating implicitly with respect to t, we have}\\ \frac{d}{dt}(2x^2-8xp+50p^2) = \frac{d}{dt}7400\\ 4x\frac{dx}{dt}- 8p\frac{dx}{dt}- 8x\frac{dp}{dt}+100p\frac{dp}{dt}=0-(1)\\ \text{Recall $\frac{dp}{dt}=2$}\\ \text{Therefore 1 becomes}\\ \frac{dx}{dt}= \frac{16x-200p}{4x-8p}-(2)\\ \text{Next, we want to obtain the values of x and p, at p = 10}\\ 2x^2-80x+5000=7400\\ =2x^2-80x-2400=0\\ \implies \text{x = 60 and x = -20}\\ \text{The only feasible value for x is 60, since x is demand and can't be }\\ \text{negative}\\ \text{Hence p = 10} \implies x =60\\ \text{Substituting in (2), we have that}\\ \frac{dx}{dt}= \frac{16(60)-200(10)}{4(60)-8(10)}=-6.5\\ \text{Therefore, the demand decreases with time at rate of 6.5}$