Question

Statement I If manufacturer can sell $x$ items at a price of $₹\left(5-\frac{x}{100}\right)$ each. The cost price of $x$ items is $₹\left(\frac{x}{5}+500\right)$. Then, the number of items he should sell to earn maximum profit is $240$ items.
Statement II The profit for selling $x$ items is given by $\frac{24}{5}x-\frac{x^2}{100}-300$

• A. Statement I is true, statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true, statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true, statement II is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (C)

Let $S(x)$ be the selling price of $x$ items and let $C(x)$ be the cost price of $x$ items. Then, we have
$S(x)=\left(5-\frac{x}{100}\right)x=5x-\frac{x^2}{100}$
and $C(x)=\frac{x}{5}+500$
Thus, the profit function $P(x)$ is given by
$P(x)=S(x)-C(x)=5x-\frac{x^2}{100}-\frac{x}{5}-500$
i.e., $P(x)=\frac{24}{5}x-\frac{x^2}{100}-500$
$\Rightarrow P^{\prime }(x)=\frac{24}{5}-\frac{x}{50}$
$P(x)=\frac{24}{5}x-\frac{x^2}{100}-500$
Now, $P^{\prime }(x)=0$ gives $x=240$.
Also, $P^{\prime \prime }(x)=\frac{-1}{50}$.
So, $P^{\prime \prime }(240)=\frac{-1}{50}<0$
Thus, $x=240$ is a point of maxima. Hence, the manufacturer can earn maximum profit. if he sells $240$ items.