# given the demand function q=9 - 1/8p and the total cost function C(q) = 1/3q^3 - 4.5q^2 + 12q+18, usinq differentiation, find the maximum profit Rearranging q = 9 - (1/8)*p, the price p is given by p = 8*(9-q)

where q is the number of items sold.

The total sales income on q items sold is q*p = 8q*(9-q)

Therefore the profit R(q) on q items sold is total sales(q) -cost(q) = q*p - C(q)

So `R(q) = 8q(9-q) - 1/3q^3 + 4.5q^2 - 12q - 18`

` ` ` = -1/3q^3 + 4.5q^2 - 12q - 18 +72q - 8q^2`

`= -1/3q^3 +(4.5-8)q^2 + (72-12)q - 18 = -1/3q^3 - 3.5q^2 +60q - 18`

To find the turning points on the profit curve R, we differentiate R and set to zero.

`d/(dq)R(q) = -3(1/3)q^2 + 2(-3.5)q + 60 = -q^2 - 7q + 60`

Therefore the turning points on the profit curve R are ` `where

`-(q^2 + 7q -60) = 0`

We could use the quadratic formula at this point to solve this, but by inspection you might see that

`-(q-5)(q+12) = 0`

because 12-5 = 7 and -5 x 12 = -60

This means that the turning points are at q=5 and q=-12. The number of items sold must be positive, so we will only investigate the turning point at q=5.

Is it a maximum or a miminum? Check the second derivative

Differentiating again we get

`d^2/(dq^2)R(q) = -2q - 7`

At q=5, this is equal to -17 which is negative, meaning that the gradient is decreasing at this point so the turning point is indeed a maximum.

The profit when q=5 items are sold is

`R(5) = -1/3 5^3 -(3.5)5^2 + 60(5) -18 = 152.83`

Maximum profit 152.83

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