If the total cost function is given by \( C(x)=\displaystyle\frac{x^3}{3}-2x^2+9x-70 \) and the selling price per unit is Rs. 6, then for what value of \( x \) will the profit be maximum?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( x=3 \)
Revenue R(x) = 6x, so the profit function is
\( \displaystyle P(x)=6x-\left(\frac{x^3}{3}-2x^2+9x-70\right) \)
\( \displaystyle =-\frac{x^3}{3}+2x^2-3x+70 \)
For a maximum, P'(x) = 0:
\( \displaystyle P'(x)=-x^2+4x-3 \)
\( \displaystyle =-(x-1)(x-3) \)
\( \displaystyle P'(x)=0\ \Rightarrow\ x=1 \text{ or } x=3 \)
Second derivative test, P''(x) = −2x + 4:
\( \displaystyle P''(1)=2\gt 0\ \text{(minimum)} \)
\( \displaystyle P''(3)=-2\lt 0\ \text{(maximum)} \)
So the profit is maximum at x = 3.