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The total cost function is given by c(x)=13x3−5x2+30x−15c(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:

Solution

✅ Correct Option: 2

The profit function is Revenue - Cost.

Given:

  • Cost: c(x)=13x3−5x2+30x−15c(x) = \frac{1}{3}x^3 - 5x^2 + 30x - 15
  • Selling price per unit = Rs. 6
  • Revenue = 6x6x

P(x)=6x−(13x3−5x2+30x−15)P(x) = 6x - \left(\frac{1}{3}x^3 - 5x^2 + 30x - 15\right)

P(x)=6x−13x3+5x2−30x+15P(x) = 6x - \frac{1}{3}x^3 + 5x^2 - 30x + 15

P(x)=−13x3+5x2−24x+15P(x) = -\frac{1}{3}x^3 + 5x^2 - 24x + 15


To find maximum profit, find the critical points:

P′(x)=−x2+10x−24P'(x) = -x^2 + 10x - 24

Setting P′(x)=0P'(x) = 0:

−x2+10x−24=0-x^2 + 10x - 24 = 0

x2−10x+24=0x^2 - 10x + 24 = 0

(x−4)(x−6)=0(x - 4)(x - 6) = 0

x=4x = 4 or x=6x = 6


To determine which gives maximum profit, use the second derivative test:

P′′(x)=−2x+10P''(x) = -2x + 10

At x=4x = 4:

P′′(4)=−2(4)+10P''(4) = -2(4) + 10

P′′(4)=2P''(4) = 2 (positive, so minimum)

At x=6x = 6:

P′′(6)=−2(6)+10P''(6) = -2(6) + 10

P′′(6)=−2P''(6) = -2 (negative, so maximum)

Therefore, profit is maximum at x=6x = 6.

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