# Price (p) and demand (x) are related by the equation: x^2+2xp+25p^2=74500(A) If the price is increasing at a rate of $2 per month when the price is $30, find the rate of change of the demand. (B)...

Price (p) and demand (x) are related by the equation: x^2+2xp+25p^2=74500

(A) If the price is increasing at a rate of $2 per month when the price is $30, find the rate of change of the demand.

(B) If the demand is decreasing at a rate of 6 units per month when the demand is 150 units, find the rate of change of the price.

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### 1 Answer

The price p and demand x are related by `x^2+2xp+25p^2=74500` . The derivative with respect to t is given by:

`2x*(dx)/(dt) + 2x*(dp)/(dt) + 2p*(dx)/(dt) + 50*p*(dp)/(dt) = 0` ...(1)

A. The rate of change of price `(dp)/(dt) = 2` when p = 30. When p = 30, `x^2 + 60x + 22500 = 74500` . Solving for x gives the roots x = 200 and x = -260. As demand cannot be negative the root x = -260 can be eliminated. Substituting the values in (1) gives:

`2*200(dx)/(dt) + 2*200*2 + 2*30*(dx)/(dt) + 50*30*2 = 0`

=> `(dx)/(dt) = -3800/460`

**The rate of change of demand is `-190/23` per month**

B. The rate of change of demand when demand is x = 150 is `(dx)/(dt) = -6`

For x = 150, `22500+300p+25p^2=74500` . Solving the equation gives p = 40 and p = -52. As price cannot be negative the root p = -52 can be eliminated.

Substituting the values in (1) gives:

`300*(-6) + 300*(dp)/(dt) + 80*(-6) + 2000*(dp)/(dt) = 0`

=> `(dp)/(dt) = 2280/2300`

**The rate of change of price is** `114/115` **per month.**