. A one-product firm estimates that its daily total cost function (in suitable units) is C (x) = x^3 - 6x^2 + 13x + 15 and its total revenue function is R(x) = 28x. Find the value of x that maximizes the daily profit. (4 points)

Value of x maximising profit : x = 5

Explanation:

Cost : C(x) = x^3 - 6x^2 + 13x + 15 ; Revenue: R(x) = 28x

Profit : Revenue - Cost = R(x) - C(x)

28x - [x^3 - 6x^2 + 13x + 15]  = 28x - x^3 + 6x^2 - 13x - 15

= - x^3 +  6x^2 + 15x - 15

To find value of 'x' that maximises total profit , we differentiate total profit function with respect to x & find that x value.

dTP/dx = - 3x^2 + 12x + 15 = 0 ► 3x^2 - 12x - 15 = 0

3x^2 + 3x - 15x - 15 = 0  ► 3x (x +1) - 15 (x + 1) = 0 ► (x+1) (3x-15) = 0

x + 1 = 0 ∴ x = -1   [Rejected, production quantity cant be negative] ;

3x - 15 = 0 ∴ 3x = 15 ∴ x = 15/3 = 5

Double derivate : d^2TP/dx^2 = - 6x + 12

d^2TP/dx^2  i.e - 6x + 12  at x = 5 is -6(5) + 12 = - 30+ 12 = -8 which is negative. So profit function is maximum at x = 5

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