The total cost function is given by \(C(x) = \displaystyle\frac{1}{3}x^3-5x^2 + 30x - 15\) and selling price per unit is Rs. 6. The profit is maximum if the value of \(x\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x = 6\)
The revenue is \(R(x) = 6x\), so the profit function is
\( \displaystyle P(x) = 6x - \left(\frac{1}{3}x^3 - 5x^2 + 30x - 15\right) \)
\( \displaystyle = -\frac{1}{3}x^3 + 5x^2 - 24x + 15 \)
Setting the first derivative to zero:
\(P'(x) = -x^2 + 10x - 24\)
\(-(x - 4)(x - 6) = 0\)
\(x = 4 \text{ or } x = 6\)
By the second derivative test:
\(P''(x) = -2x + 10\)
\(P''(4) = 2 \gt 0\) (minimum)
\(P''(6) = -2 \lt 0\) (maximum)
So the profit is maximum at \(x = 6\).
Why the other options are incorrect
Option (1) \(x = 4\) — Incorrect. \(P'(4) = 0\), but \(P''(4) = 2 \gt 0\), so the profit is a minimum there.
Option (3) \(x =5\) — Incorrect. \(P'(5) = -25 + 50 - 24 = 1 \ne 0\).
Option (4) \(x = 10\) — Incorrect. \(P'(10) = -100 + 100 - 24 = -24 \ne 0\).