Answer:
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
