Question

The total cost function of a firm is given by `C(x) = 1/3 x^3 - 5x^2 + 30x - 15` where the selling price per unit is given as ₹ 6. Find for what value of x will the profit be maximum.

Answer 1

Given: `C(x) = 1/3 x^3 - 5x^2 + 30x - 15`

∴ P(x) = S(x) – C(x)   ...[∵ P = 6 ∴ S(x) = 6x]

= `6x - 1/3 x^3 + 5x^2 - 30x + 15`

= `5x^2 - 1/3 x^3 - 24x + 15`

Now, `(dP)/dx = 10x - (3x^2)/3 - 24`

= 10x – x² – 24

For maximum and minimum profit,

Put `(dP)/dx = 0`

⇒ 10x – x² – 24 = 0

⇒ x² – 10x + 24 = 0

⇒ (x – 6) (x – 4) = 0

⇒ x = 6 or x = 4

Now, `(d^2P)/(dx^2) = 10 - 2x`

∴ `((d^2P)/(dx^2))_(x = 6) = 10 - 2 xx 6`

= –2 < 0

So, at x = 6, P(x) is maximum.

And `((d^2P)/(dx^2))_(x = 4) = 10 - 2 xx 4`

= 2 > 0

So, at x = 4, P(x) is minimum.

Hence, at x = 6, profit is maximum.