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(A) Given,cost function $C(x) = x^2 + 78x + 2500$. Price per unit $p = 600 - 8x$. Revenue function $R(x) = x 	imes p = x(600 - 8x) = 600x - 8x^2$. Pr

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Rs. $5069$

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(A) Given,cost function $C(x) = x^2 + 78x + 2500$. Price per unit $p = 600 - 8x$. Revenue function $R(x) = x 	imes p = x(600 - 8x) = 600x - 8x^2$. Profit function $P(x) = R(x) - C(x) = (600x - 8x^2) - (x^2 + 78x + 2500) = -9x^2 + 522x - 2500$. To find maximum profit,find the derivative $P'(x)$ and set it to $0$: $P'(x) = -18x + 522 = 0 \implies 18x = 522 \implies x = 29$. Check the second derivative: $P''(x) = -18 Maximum profit $P(29) = -9(29)^2 + 522(29) - 2500 = -9(841) + 15138 - 2500 = -7569 + 15138 - 2500 = 5069$. Thus,the maximum profit is Rs. $5069$.

$A$ manufacturer produces $x$ items per week at a total cost of Rs $(x^2+78x+2500)$. The price per unit is given by $8x = 600 - p$,where $p$ is the price of each unit. Then the maximum profit obtained is

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