The total cost function is given by $C(x) =\frac{1}{3}x^3-5x^2 + 30x – 15$ and the selling price per unit is ₹6. Find for what value of x will the profit be maximum.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $x=6$
Since the selling price per unit is ₹6,
∴ revenue function $R(x) = 6x$.
Given total cost function $C(x) = \frac{1}{3}x^3 - 5x^2+30x - 15$.
Profit function $P(x)$ = Revenue - Cost = $R(x)-C(x)$
$⇒P(x) = 6x-\left(\frac{1}{3}x^3 - 5x^2+30x - 15\right)=-\frac{1}{3}x^3+5x^2-24x+15$
To find the value(s) of x when P(x) is maximum, we should find value(s) of x when $\frac{dP}{dx}=0$ and $\frac{d^2P}{dx^2}<0$.
Now, $\frac{dP}{dx}=-\frac{1}{3}.3x^2+5.2x-24.1+0=-x^2+10x - 24$ and $\frac{d^2P}{dx^2}= -2x + 10$.
$\frac{dP}{dx}=0⇒-x^2+10x-24=0⇒ x^2 - 10x + 24 = 0$
$⇒(x-4)(x-6)=0⇒x=4, 6$.
At $x = 4,\frac{d^2P}{dx^2}=-2 × 4+10=2> 0$ and
at $x = 6,\frac{d^2P}{dx^2}=-2 × 6+10=-2<0$.
Hence, the profit is maximum when $x = 6$.