Answer to Question #265671 in Calculus for ozan.ture

Question #265671

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.


1
Expert's answer
2021-11-15T16:40:26-0500

"\\displaystyle\n2x^2 -8xp+2x^2=7400, \\frac{dp}{dt} = 2\\\\\n\\text{Differentiating implicitly with respect to t, we have}\\\\\n\\frac{d}{dt}(2x^2-8xp+50p^2) = \\frac{d}{dt}7400\\\\\n4x\\frac{dx}{dt}- 8p\\frac{dx}{dt}- 8x\\frac{dp}{dt}+100p\\frac{dp}{dt}=0-(1)\\\\\n\\text{Recall $\\frac{dp}{dt}=2$}\\\\\n\\text{Therefore 1 becomes}\\\\\n\\frac{dx}{dt}= \\frac{16x-200p}{4x-8p}-(2)\\\\\n\\text{Next, we want to obtain the values of x and p, at p = 10}\\\\\n2x^2-80x+5000=7400\\\\\n=2x^2-80x-2400=0\\\\\n\\implies \\text{x = 60 and x = -20}\\\\\n\\text{The only feasible value for x is 60, since x is demand and can't be }\\\\\n\\text{negative}\\\\\n\\text{Hence p = 10} \\implies x =60\\\\\n\\text{Substituting in (2), we have that}\\\\\n\\frac{dx}{dt}= \\frac{16(60)-200(10)}{4(60)-8(10)}=-6.5\\\\\n\\text{Therefore, the demand decreases with time at rate of 6.5}"


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