The cost of making $x$ units is $C(x) = x^{3} - 9x^{2} + 40x + 100$ rupees. At what output is the marginal cost least?

3 unitsCorrect answerYour answer4 unitsCorrect answerYour answer6 unitsCorrect answerYour answer9 unitsCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$MC = 3x^{2} - 18x + 40$ . It is least where its own derivative is 0: $6x - 18 = 0$ .
$x = 3$ units, and the second derivative 6 is positive, so it is a minimum.

Worth noting

9 units halves 18 but forgets that the square term is $3x^{2}$ : the lowest point of $3x^{2} - 18x + 40$ is at $\frac{18}{2 \times 3} = 3$ .

Ten questions from Rate of Change, Marginal Cost & Revenue, at your own pace.

Ten questions from Rate of Change, Marginal Cost & Revenue, at your own pace.

The cost of making xxx units is C(x)=x3−9x2+40x+100C(x) = x^{3} - 9x^{2} + 40x + 100C(x)=x3−9x2+40x+100 rupees. At what output is the marginal cost least?