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Question

(A) Let $S(x)$ be the total revenue from selling $x$ items and $C(x)$ be the cost price of $x$ items. Then,$S(x) = x \left(5 - \frac{x}{100}\right) =

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(A) Let $S(x)$ be the total revenue from selling $x$ items and $C(x)$ be the cost price of $x$ items. Then,$S(x) = x \left(5 - \frac{x}{100}ight) = 5x - \frac{x^2}{100}$. The cost function is $C(x) = \frac{x}{5} + 500$. The profit function $P(x)$ is defined as $P(x) = S(x) - C(x)$. $P(x) = \left(5x - \frac{x^2}{100}ight) - \left(\frac{x}{5} + 500ight) = 5x - \frac{x^2}{100} - 0.2x - 500 = 4.8x - \frac{x^2}{100} - 500$. To find the maximum profit,we find the derivative $P'(x)$ and set it to zero: $P'(x) = 4.8 - \frac{2x}{100} = 4.8 - \frac{x}{50}$. Setting $P'(x) = 0$,we get $4.8 = \frac{x}{50}$,which implies $x = 4.8 	imes 50 = 240$. To verify it is a maximum,we check the second derivative $P''(x) = -\frac{1}{50}$. Since $P''(240) = -\frac{1}{50} < 0$,the profit is maximum at $x = 240$.

$A$ manufacturer can sell $x$ items at a price of rupees $\left(5 - \frac{x}{100}\right)$ each. The cost price of $x$ items is $\text{Rs} \left(\frac{x}{5} + 500\right)$. Find the number of items he should sell to earn maximum profit.

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