Question

Find the output yielding maximum profit for the cost function
C =0.7 x^3 − 0.8 x^2 + 12x +9
given that the cost price of x is Rs 45/- per unit.

Answer 1

Solution. We write the profit function P(x)=R(x)-C(x).

$$P(x)=45x-(0.7x^3-0.8x^2+12x+9)=-0.7x^3+0.8x^2+33x-9$$

Find the value of x for which the function P(x) takes the maximum value. Find the derivative P (x)

$$P'(x)=-2.1x^2+1.6x+33$$

Find the points at which the derivative is zero.

$$-2.1x^2+1.6x+33=0$$

$$D=1.6^2-4(-2.1)\times33=279.76$$

Roots of the equation

$$x_1=\frac {-1.6-\sqrt{279.76}}{-2\times2.1}\approx 4.36$$

$$x_2=\frac {-1.6+\sqrt{279.76}}{-2\times2.1}\approx -3.6<0$$

Find the value of the derivative in each interval (considering that x>0)

1. 0<x<x₂ P'(x)>0 function P (x) increases.
2. x>x₂ P'(x)>0 function P (x) decreases.

As result maximum profit for the cost function P(x)=92=P(4.36).

Answer. 92.