Question

A manufacturer can sell $x$ items at the rate of $(330-x)$ each. The cost of producing items is $Rs.(x^2+10x-12)$. How many items must be sold so that his profit is maximum.

• A. $80$
• B. $20$
• C. $70$
• D. $50$

Answer 1

The correct option is A $80$

Selling price of $1$ item $=Rs.(330-x)\left(\text{Given}\right)$

Selling price of $x$ items $=Rs.x(330-x)$

Cost price of $x$ items $=Rs.(x^2+10x-12)$

Profit $=$ Selling price $-$ Cost price

$\Rightarrow$ Profit $=(330x-x^2)-(x^2+10x-12)=320x-2x^2+12$

Now,

$P=-2x^2+320x+12$

Differentiating above equation, we get

$\frac{dP}{dx}=-4x+320$

Differentiating againg w.r.t. $x$, we get

$\frac{d^{2}P}{d{x}^2}=-4\Rightarrow$ Always negative

To maximize the profit,

$\frac{dp}{dx}=0-4x+320=0x=\frac{-320}{-4}=80$

$\because \frac{d^{2}P}{d{x}^2}<0$ for any value of $x$, thus $80$ items must be sold so that his profit is maximum.