A firm has the cost function; $f(x)=\frac{x^3}{3}-7 x^2+111 x+50$ and demand function is $x=100-p$; based on this information the value of x, when profit is maximum, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11
$x = 100 - p \Rightarrow p = 100 - x$
$\text{Revenue } R = px = x(100 - x) = 100x - x^2$
$\text{Cost } C = \frac{x^3}{3} - 7x^2 + 111x + 50$
$\text{Profit } P = R - C$
$= 100x - x^2 - \left(\frac{x^3}{3} - 7x^2 + 111x + 50\right)$
$= -\frac{x^3}{3} + 6x^2 - 11x - 50$
$P' = -x^2 + 12x - 11$
$-x^2 + 12x - 11 = 0$
$x^2 - 12x + 11 = 0$
$x = 1,\; 11$
$P'' = -2x + 12$
$P''(1)=10>0 \Rightarrow \text{min},\;\; P''(11)=-10<0 \Rightarrow \text{max}$
The value of $x$ for maximum profit is $11$.