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A price p (in dollars) and demand x for a product are related by

2x^2−10xp+50p^2=6200.


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.


Rate of change of demand = ?

Verified Answer :

Plugging in p = 10 to solve for x:

x^2-50x-600 = (x-60)(x+10) = 0 => x = 60 
x^2 - 5xp + 50p^2 = 3100 

Differentiate with respect to time, 

2xx' - 50x'p - 50xp' + 100pp' = 0 
240 - 100(10) - 3000p' + 1000p' = 0 

p' = -.38


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Using implicit differentiation, it is found that the rate of change of demand is of 21.64.

The equation that describes the relationship between price p and demand x is given by:

\(2x^2 - 10xp + 50p^2 = 6200\)

In this problem, we have that the price is \(p = 10\), thus, the demand is found solving the equation:

\(2x^2 - 100x + 5000 = 6200\)

\(2x^2 - 100x - 1200 = 0\)

\(x^2 - 10x - 120 = 0\)

Which is a quadratic equation with \(a = 1, b = 10, c = -120\). Then:

\(\Delta = 10^2 - 4(1)(-120) = 580\)

\(x_{1} = \frac{-10 + \sqrt{580}}{2} = 7.04\)

\(x_{2} = \frac{-10 - \sqrt{580}}{2} = -17.04\)  

Since the demand is positive, we use \(x = 7.04\).

The rate of change, applying implicit differentiation, is given by:

\(2x\frac{dx}{dt} - 10p\frac{dx}{dt} - 10x\frac{dp}{dt} + 100p\frac{dp}{dt} = 0\)

We have that \(\frac{dp}{dt} = 2\), thus:

\(2(7.04)\frac{dx}{dt} - 100\frac{dx}{dt} - 10(7.04)(2) + 100(10)(2) = 0\)

\(\frac{dx}{dt}= \frac{10(7.04)(2) - 100(10)(2)}{2(7.04) - 100}\)

\(\frac{dx}{dt} = 21.64\)

The rate of change of demand is of 21.64.

A similar problem is given at https://brainly.com/question/25081524

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