Class 12 Maths, Worked in Full Past papersGet updates
Chapter 6 · Application of Derivatives

Example 5: Marginal cost of 0.005x3−0.02x2+30x+50000.005x^3-0.02x^2+30x+5000 at x=3x = 3

ExamplesRate of change of quantities · Section 6.2
The question

The total cost C(x)\mathrm{C}(x) in Rupees, associated with the production of xx units of an item is given by

C(x)=0.005 x3−0.02 x2+30x+5000\mathrm{C}(x) = 0.005\,x^3 - 0.02\,x^2 + 30x + 5000

Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

Exam answer

Marginal cost MC=dCdx\mathrm{MC} = \frac{d\mathrm{C}}{dx}⁠. So

MC=0.015x2−0.04x+30\mathrm{MC} = 0.015x^2 - 0.04x + 30

When x=3x = 3⁠,

MC=0.015(9)−0.04(3)+30=0.135−0.12+30=30.015\begin{aligned} \mathrm{MC} &= 0.015(9) - 0.04(3) + 30 \\ &= 0.135 - 0.12 + 30 \\ &= 30.015 \end{aligned}

Hence the required marginal cost is ₹30.02 (nearly).

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

"Instantaneous rate of change of total cost" with respect to output is the derivative C′(x)\mathrm{C}'(x)⁠. So marginal cost is just the derivative of the cost function, evaluated at the output level asked for. In practice it is close to the extra cost of making one more unit.

Given, to find, and the plan
Given
C(x)=0.005x3−0.02x2\mathrm{C}(x) = 0.005x^3 - 0.02x^2 + 30x+5000+\, 30x + 5000 rupees for xx units.
To find
The marginal cost MC=dCdx\mathrm{MC} = \dfrac{d\mathrm{C}}{dx} at x=3x = 3⁠.
Plan
Differentiate term by term; substitute x=3x = 3 carefully; round to paise; compare with the actual cost of the next unit.

Basics you need

1A rate of change is a derivative

The rate of change of C\mathrm{C} with respect to xx is dCdx\frac{d\mathrm{C}}{dx} (Section 6.2). Here it is the marginal cost, measured in rupees per unit. Basics: rate of change as a derivative

2Differentiating a polynomial

Power rule ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}⁠, constant multiple rule, sum rule; the derivative of a constant is 0 (Table 5.3 and the algebra of derivatives, Section 5.3).

3Decimal arithmetic

0.005×3=0.0150.005 \times 3 = 0.015⁠, 0.02×2=0.040.02 \times 2 = 0.04⁠; 0.015×9=0.1350.015 \times 9 = 0.135 and 0.04×3=0.120.04 \times 3 = 0.12⁠.

Solution

  1. Differentiate the cost function

    Apply the power rule to each term, keeping each constant multiplier; the fixed cost 5000 has derivative 0:

    MC=dCdx=0.005(3x2)−0.02(2x)+30=0.015x2−0.04x+30.\begin{aligned} \mathrm{MC} = \frac{d\mathrm{C}}{dx} &= 0.005(3x^2) - 0.02(2x) + 30 \\ &= 0.015x^2 - 0.04x + 30 . \end{aligned}
    The definition in the question ("instantaneous rate of change of total cost") is exactly what a derivative measures. The constant 5000 is a fixed cost, paid whatever the output; it does not change as xx changes, so it contributes nothing to the rate.
    Term by term: ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2⁠, ddx(x2)=2x\frac{d}{dx}(x^2) = 2x⁠, ddx(x)=1\frac{d}{dx}(x) = 1⁠, ddx(5000)=0\frac{d}{dx}(5000) = 0⁠. Basics: rate of change as a derivative

  2. Substitute x=3x = 3

    MC(3)=0.015(32)−0.04(3)+30=0.135−0.12+30=30.015.\begin{aligned} \mathrm{MC}(3) &= 0.015(3^2) - 0.04(3) + 30 \\ &= 0.135 - 0.12 + 30 \\ &= 30.015 . \end{aligned}

    In money, 30.015 rupees is 30 rupees and 1.5 paise, so to the nearest paisa the marginal cost is about ₹30.02 per unit.

    Square first: 32=93^2 = 9⁠, then 0.015×9=0.1350.015 \times 9 = 0.135⁠; next 0.04×3=0.120.04 \times 3 = 0.12⁠. Then 0.135−0.12=0.0150.135 - 0.12 = 0.015⁠, and adding 30 gives 30.015. The book rounds the half paisa up, writing "₹30.02 (nearly)".
    Units: C\mathrm{C} in rupees, xx in units of the item, so MC\mathrm{MC} is in rupees per unit.

  3. What the number means

    Compare with actual costs. Exactly,

    C(2)=5059.96,C(3)=5089.955,C(4)=5120.\begin{aligned} \mathrm{C}(2) &= 5059.96, \\ \mathrm{C}(3) &= 5089.955, \\ \mathrm{C}(4) &= 5120 . \end{aligned}

    So the 3rd unit costs C(3)−C(2)=29.995\mathrm{C}(3) - \mathrm{C}(2) = 29.995 rupees and the 4th unit costs C(4)−C(3)=30.045\mathrm{C}(4) - \mathrm{C}(3) = 30.045 rupees. The marginal cost 30.015 at x=3x = 3 lies between them; their average is C(4)−C(2)2=30.02\frac{\mathrm{C}(4) - \mathrm{C}(2)}{2} = 30.02⁠.

    Over one unit the cost function bends only slightly, so the derivative at 3 is a good estimate of the cost of one more unit at that level of output. This is why economists use the marginal cost as "the cost of producing one more unit".
    In general C(3+h)−C(3)=30.015h+0.025h2+0.005h3.\begin{aligned} &\mathrm{C}(3+h) - \mathrm{C}(3) \\ &\quad = 30.015h + 0.025h^2 + 0.005h^3 . \end{aligned} For small hh the first term dominates, and C(3+h)−C(3)h→30.015\frac{\mathrm{C}(3+h) - \mathrm{C}(3)}{h} \to 30.015 as h→0h \to 0⁠.
Result

MC=C′(x)=0.015x2−0.04x+30\mathrm{MC} = \mathrm{C}'(x) = 0.015x^2 - 0.04x + 30⁠, so when 3 units are produced

MC(3)=30.015≈30.02 rupees per unit.\mathrm{MC}(3) = 30.015 \approx 30.02\ \text{rupees per unit}.

This agrees with the book's solution (₹30.02, nearly).

Check it with numbers

C(3)=0.005(27)−0.02(9)\mathrm{C}(3) = 0.005(27) - 0.02(9) + 90+5000+\, 90 + 5000 =0.135−0.18+5090= 0.135 - 0.18 + 5090 =5089.955= 5089.955⁠. With a small step h=0.001h = 0.001⁠,

C(3.001)−C(3)0.001≈30.015025,\frac{\mathrm{C}(3.001) - \mathrm{C}(3)}{0.001} \approx 30.015025 ,

which agrees with 30.015 to four decimal places (the extra 0.000025 is the 0.025h0.025h term, and it shrinks to 0 with hh⁠).

Common mistakes
  1. Giving the total cost C(3)=5089.955\mathrm{C}(3) = 5089.955 instead of its rate of change.
  2. Keeping the 5000 in the derivative. A constant has derivative 0.
  3. Decimal slips: 0.005×3=0.0150.005 \times 3 = 0.015⁠, not 0.15; 0.02×2=0.040.02 \times 2 = 0.04⁠.
  4. Dividing C(3)\mathrm{C}(3) by 3. 5089.9553≈1696.65\frac{5089.955}{3} \approx 1696.65 is the average cost per unit, a different idea.
What to take away
  1. Marginal cost = dCdx\frac{d\mathrm{C}}{dx}⁠, in rupees per unit.
  2. It approximates the cost of one more unit at that output level.
  3. Fixed costs do not affect marginal cost.

Every fact used, and where it comes from

FactUsed inSource
Marginal cost is the instantaneous rate of change of total costStep 1Stated in the question; Section 6.2 (Lesson: rate of change of quantities)
Power, constant multiple and sum rules; derivative of a constant is 0Step 1Table 5.3 and the algebra of derivatives, Section 5.3
Rate of change as a derivativeSteps 1, 3Section 6.2 (Basics: rate of change)
Values 30.015, 5059.96, 5089.955, 5120, 29.995, 30.045, 30.015025Steps 2, 3; CheckArithmetic, recomputed in the site's verification script
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →