Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.7 Junior Find the answer

The sales revenue $y_1$ (in ten thousand yuan) of machines produced by a certain factory is a function of the production volume $x$ (in thousand units): $y_1 = 17x^2$ . The total production cost $y_2$ (in ten thousand yuan) is also a function of the production volume $x$ (in thousand units): $y_2 = 2x^3 - x^2$ ( $x > 0$ ). To maximize profit, the production should be $\quad$ .

Pick one

Solution

Solution: The profit $y = y_1 - y_2 = 17x^2 - (2x^3 - x^2) = 18x^2 - 2x^3$ . The derivative of profit with respect to $x$ is $y' = -6x^2 + 36x$ .

Solving $y' > 0$ yields $0 6$ .

When $x = 6$ , $y$ reaches its maximum value.

Therefore, the answer is $\boxed{A}$ .

By calculating profit as revenue minus cost, we find $y = y_1 - y_2$ and discuss the sign of $y'$ to determine the maximum value of the function.

This tests the student's ability to use derivatives to find the maximum value of a function on a closed interval.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.