MHT CET · Maths · Application of Derivatives
A manufacturer produces \(x\) items per week at a total cost of Rs \(\left(x^2+78 x+2500\right)\). The price per unit is given by \(8 x=600-\mathrm{p}\) where ' p ' is the price of each unit. Then the maximum profit obtained is
- A Rs. 5069
- B Rs. 15138
- C Rs. 7569
- D Rs. 2500
Answer & Solution
Correct Answer
(A) Rs. 5069
Step-by-step Solution
Detailed explanation
Profit: \(P(x) = x(600-8x) - (x^2+78x+2500) = -9x^2 + 522x - 2500\) For max profit: \(P'(x) = -18x + 522 = 0 \implies x = 29\)
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