CA Foundation · Quantitative Aptitude · Differential and Integral Calculus
A manufacturing company's profit function is given by P(x) = -2x² + 80x - 300, where x is the number of units produced (in hundreds). At what production level is the profit maximized?
The profit function P(x) = -2x² + 80x - 300 is maximized by finding where P'(x) = -4x + 80 = 0, which gives x = 20 hundred units. The second derivative P''(x) = -4 is negative, confirming a maximum.
- A10 hundred units
- B20 hundred unitsCorrect
- C30 hundred units
- D40 hundred units
Explanation
To find the maximum profit, we take the derivative and set it equal to zero: P'(x) = -4x + 80 = 0. Solving for x gives x = 20 hundred units. We confirm this is a maximum by checking the second derivative: P''(x) = -4 < 0, confirming a maximum. A distractor of 10 units results from solving -2x + 40 = 0, an arithmetic error in differentiation.
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