A firm has the cost function $C =\frac{x^3}{3}-7x^2 + 11x+50$ and demand function $x = 100 – p$. Write the total revenue function in terms of x. Formulate the total profit function P in terms of x. Find the profit maximizing level of output x and what is the maximum profit?

$x=11$, Max Profit = ₹111.33

The correct answer is Option (3) → $x=11$, Max Profit = ₹111.33

Demand function is $x = 100-p$

$⇒p=100-x$.

∴ Revenue function, $R(x) = p.x = x(100-x) = 100x-x^2$.

Profit function, $P(x)$ = Revenue - Cost = $R(x)-C(x)$.

$⇒ P(x) = (100x – x^2) – \left(\frac{x^3}{3}-7x^2 + 11x+50\right) = -\frac{x^3}{3}+6x^2-11x-50$   ...(1)

To find values of x so that P(x) is maximum, we should find value(s) of x where $\frac{dP}{dx}= 0$ and $\frac{d^2p}{dx^2}< 0$.

Now $\frac{dP}{dx}=0⇒-x^2 + 12x-11=0⇒x^2-12x + 11 = 0$

$⇒ (x-1) (x-11)=0⇒x=1, 11$.

$\frac{d^2p}{dx^2}=\frac{d}{dx}(-x^2 + 12x-11) = -2x+12$.

At $x = 1,\frac{d^2p}{dx^2}=-2 × 1+12=10 > 0$ and

at $x = 11,\frac{d^2p}{dx^2}=-2 × 11+12=-10 < 0$.

Hence, the profit maximising level is at $x = 11$.

Putting $x = 11$ in (1), we get

$P(11)=-\frac{(11)^3}{3}+6 × (11)^2 - 11 × 11 - 50 =-\frac{1331}{3}+726-121-50$

$=-\frac{1331}{3}+555=111.33$

Hence, maximum profit = ₹111.33