A manufacturer sells x items at a price of ru | vedclass

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(C) Let $R(x)$ be the revenue function and $C(x)$ be the cost function.Revenue $R(x) = x 	imes \left(6 - \frac{x}{40}ight) = 6x - \frac{x^2}{40}$.C

Vedclass · Accepted Answer

$143.4$

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(C) Let $R(x)$ be the revenue function and $C(x)$ be the cost function.Revenue $R(x) = x 	imes \left(6 - \frac{x}{40}ight) = 6x - \frac{x^2}{40}$.Cost $C(x) = \frac{x}{5} + 193$.Profit $P(x) = R(x) - C(x) = 6x - \frac{x^2}{40} - \frac{x}{5} - 193$.$P(x) = -\frac{x^2}{40} + \frac{29x}{5} - 193$.To find the maximum profit,we find the derivative $P'(x)$ and set it to $0$.$P'(x) = -\frac{2x}{40} + \frac{29}{5} = -\frac{x}{20} + 5.8$.Setting $P'(x) = 0$,we get $\frac{x}{20} = 5.8$,so $x = 116$.Now,find the second derivative $P''(x) = -\frac{1}{20}$. Since $P''(x) Maximum profit $P(116) = -\frac{116^2}{40} + \frac{29(116)}{5} - 193$.$P(116) = -\frac{13456}{40} + 672.8 - 193 = -336.4 + 672.8 - 193 = 143.4$.

$A$ manufacturer sells $x$ items at a price of rupees $\left(6-\frac{x}{40}\right)$ each. The cost price of $x$ items is Rs. $\left(\frac{x}{5}+193\right)$. The maximum profit in Rs. is

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