The price function is p = 8(9-q)

The sales function is price x quantity sold = pq = 8q(9-q) = S(q)

The profit function is given by

R = S(q) - C(q)

where C(q) is the cost function

If the cost function is given by

C(q) = 1/3q^3 - 4.5q^2 + 12q + 18

then

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

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

Gathering terms

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

= -1/3q^3 - 3.5q^2 + 60q - 18

To find the maxima and minima of R(q), differentiate and set to zero:

ie, solve R'(q) = 0

Now, R'(q) = -1/3(3)q^2 - (3.5)2q + 60

Setting this equal to zero gives

-q^2 - 7q + 60 = 0

Multiply through by -1 giving

q^2 + 7q - 60 = 0

Noticing that -5 x 12 = -60 and -5 + 12 = 7, this factorises to

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

`therefore` the maxima and minima of R(q) are at q = 5 and q = -12

The quantity of items sold must be positive, so we only consider the root q =5

To check that the turning point at q = 5 is a maximum, differentiate R(q) again

R''(q) = d/(dq) R'(q) = -2q - 7

when q = 5, this is negative showing that the gradient of the profit R(q) is decreasing at that point. This result is consistent with the maximum profit being achieved at q = 5.

At q = 5, the profit is given by R(5) = -1/3(5)^3 -3.5(5)^2 + 60(5) - 18 = 152.83

**maximum profit is 152.83**

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