The total cost function is given by $C(x) = \frac{1}{3}x^3-5x^2 + 30x - 15$ 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) → $x = 6$ **
$C(x)=\frac{1}{3}x^{3}-5x^{2}+30x-15$
$R(x)=6x$
$P(x)=R(x)-C(x)=6x-\left(\frac{1}{3}x^{3}-5x^{2}+30x-15\right)=-\frac{1}{3}x^{3}+5x^{2}-24x+15$
$P'(x)=-x^{2}+10x-24$
Solve $P'(x)=0$: $-x^{2}+10x-24=0\Rightarrow x^{2}-10x+24=0$
Discriminant $D=10^{2}-4\cdot1\cdot24=100-96=4$
$x=\frac{10\pm\sqrt{4}}{2}=\frac{10\pm 2}{2}\Rightarrow x=4,\ 6$
$P''(x)=-2x+10$
$P''(4)=2>0$ (local minimum), \ $P''(6)=-2<0$ (local maximum)
$x=6$