Question

A manufacturer can sell x items at a price of Rs.(5-x/100) each. The cost price of x items is Rs.(x/5+500).

Answer 1

So, Profit is maximum at x=240

He should sell 240 products to earn maximum profit.

Answer 2

Correct Question

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☞manufacture can sell X item as a price of rupees (5-x/100) each the cost of price of X item Rs (x/5 +500) find the number of items he should sell to earn maximum profit.

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\sf let \: s \: (x) \: be \: the \: selling \: price \: of \: x \: items \: and \: let \: c(x) \: e \: the \: cost \: of \: x \: items \ \\ \sf \: so \: that \: we \: have \\ \\ \sf s(x) = (5 - \frac{x}{100} )x = 5x - \frac{ {x}^{2} }{100} \\ \\ \sf \: c(x) = \frac{x}{5} + 500 \\ \\ \sf therefore \: the \: profit \: function \: p(x)is \: given \: like \: \\ \\ \sf p(x) = s(x) - c(x) = 5x - \ \frac{ {x}^{2} }{100} - \frac{x}{5} - 500 \\ \\ \sf p(x) = \frac{24}{5} x - \frac{ {x}^{2} }{100} - 500 \\ \\ \sf {p}^{l} = \frac{24}{5} - \frac{x}{50} \\ \\ \sf {p}^{l} (x) = 0 \: gives \: x = 240. \\ \\ \sf \: also \: {p}^{ll} (x) = \frac{ - 1}{50} . \\ \\ \sf {p}^{ll} (240) = \frac{ - 1}{50 } < 0

Thus , x=240 is a point of maxima .....

So, the manufacturer can earn maximum profit , if he sells 240 items _____