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Chapter 6 · Application of Derivatives

Exercise 6.1 · Q15: Marginal cost of C(x)C(x) at x=17x = 17 units

Exercise 6.1Rate of change of quantities · Section 6.2
The question

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

C(x)=0.007x3−0.003x2+15x+4000.C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000.

Find the marginal cost when 17 units are produced.

Exam answer

Marginal cost =C′(x)= C'(x)⁠.

C′(x)=0.021x2−0.006x+15C'(x) = 0.021x^2 - 0.006x + 15 C′(17)=0.021(17)2−0.006(17)+15=6.069−0.102+15=20.967\begin{aligned} C'(17) &= 0.021(17)^2 \\ &\quad - 0.006(17) + 15 \\ &= 6.069 - 0.102 + 15 \\ &= 20.967 \end{aligned}

∴\therefore the marginal cost is ₹20.96720.967⁠.

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

The one idea this problem is built on

Marginal cost is the instantaneous rate of change of total cost with respect to output (the book's definition in Example 5). So it is the derivative C′(x)C'(x)⁠, evaluated at x=17x = 17⁠. It is not the total cost C(17)C(17)⁠. Roughly, it tells us how much one more unit would add to the cost at that level of production.

Given, to find, and the plan
Given
C(x)=0.007x3−0.003x2+15x+4000C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000 rupees for xx units.
To find
The marginal cost MC=C′(x)\text{MC} = C'(x) at x=17x = 17⁠.
Plan
(1) Recall what marginal cost means. (2) Differentiate CC term by term. (3) Substitute x=17x = 17 with careful decimal arithmetic. (4) State the answer with units, and interpret it.

Basics you need

1Marginal cost is a derivative

Marginal cost = dCdx\frac{dC}{dx}⁠, the rate of change of total cost with respect to the number of units produced (Example 5, Section 6.2). Basics: rate of change as a derivative

2Differentiating a polynomial

Differentiate term by term: ddx(axn)=naxn−1\frac{d}{dx}\left(ax^n\right) = nax^{n-1}⁠, ddx(ax)=a\frac{d}{dx}(ax) = a⁠, and a constant gives 0.

3Decimal multiplication

0.007×3=0.0210.007\times 3 = 0.021 and 0.003×2=0.0060.003\times 2 = 0.006⁠: multiply the digits, keep the same number of decimal places.

Solution

  1. Marginal cost means C′(x)C'(x)

    By the definition used in Example 5, the marginal cost at output xx is

    MC=dCdx=C′(x).\text{MC} = \frac{dC}{dx} = C'(x).

    Its unit is rupees per unit of output.

    "Marginal" in economics means "at the margin": the rate at which cost changes as output changes. That is exactly a derivative. The total cost C(17)C(17) answers a different question (how much do 17 units cost altogether?).
    Rate of change of yy with respect to xx is dydx\frac{dy}{dx} (Section 6.2). Basics: rate of change as a derivative

  2. Differentiate term by term

    C′(x)=0.007(3x2)−0.003(2x)+15+0=0.021x2−0.006x+15.\begin{aligned} C'(x) &= 0.007(3x^2) - 0.003(2x) + 15 + 0 \\ &= 0.021x^2 - 0.006x + 15. \end{aligned}
    Each term is a constant times a power of xx⁠, so the power rule applies to each, and the derivative of a sum or difference is the sum or difference of the derivatives. The fixed cost 4000 (the cost even when nothing is produced) is a constant, so its derivative is 0: overheads do not change the cost of producing one more unit.
    ddx(xn)=nxn−1\frac{d}{dx}\left(x^n\right) = nx^{n-1} (Table 5.3) and (u±v)′=u′±v′(u \pm v)' = u' \pm v' (Section 5.3, rule (1)).

  3. Substitute x=17x = 17

    First the pieces: 172=28917^2 = 289⁠, so

    0.021×289=6.069,0.006×17=0.102.\begin{aligned} 0.021\times 289 &= 6.069, \\ 0.006\times 17 &= 0.102. \end{aligned}

    Then

    C′(17)=6.069−0.102+15=20.967.\begin{aligned} C'(17) &= 6.069 - 0.102 + 15 \\ &= 20.967. \end{aligned}
    Computing each term separately keeps the decimal places under control: 21×289=606921\times 289 = 6069⁠, and three decimal places gives 6.069; 6×17=1026\times 17 = 102⁠, and three decimal places gives 0.102.

  4. State and interpret the answer

    The marginal cost when 17 units are produced is ₹20.967 per unit (about ₹20.97).

    Interpretation: at an output of 17 units, the total cost is rising at about ₹20.97 for each extra unit. Producing the 18th unit actually adds C(18)−C(17)C(18) - C(17)⁠, which is 4309.852−4288.524=21.3284309.852 - 4288.524 = 21.328 rupees, close to (but not exactly) the marginal cost, because C′C' gives the instantaneous rate at x=17x = 17 and the rate keeps rising between 17 and 18 (its own derivative, C′′(x)=0.042x−0.006C''(x) = 0.042x - 0.006⁠, is positive for every x≥1x \ge 1⁠).

    The derivative is the rate at one level of output. The exact cost of one extra whole unit is a difference over a step of 1, which is only approximately the derivative. Example 5 in the book rounds its answer to ₹30.02 "nearly"; here the textbook keeps the exact value ₹20.967.
    Rate of change of cost with respect to output. Lesson: rate of change of quantities
Result

Marginal cost =C′(17)= C'(17)⁠, and C′(17)=0.021(17)2−0.006(17)+15C'(17) = 0.021(17)^2 - 0.006(17) + 15⁠, which is ₹20.967 per unit. (Textbook answer: ₹20.967.)

Check it with numbers

Compute the cost at outputs either side of 17:

  • C(16)=4267.904C(16) = 4267.904⁠, C(17)=4288.524C(17) = 4288.524⁠, C(18)=4309.852C(18) = 4309.852⁠.
  • The 17th unit adds C(17)−C(16)=20.620C(17) - C(16) = 20.620⁠; the 18th adds 21.32821.328⁠.
  • Their average, C(18)−C(16)2=20.974\frac{C(18) - C(16)}{2} = 20.974⁠, is very close to C′(17)=20.967C'(17) = 20.967⁠.

(For C(17)C(17)⁠: 0.007×4913=34.3910.007\times 4913 = 34.391 and 0.003×289=0.8670.003\times 289 = 0.867⁠, so C(17)=34.391−0.867+255+4000C(17) = 34.391 - 0.867 + 255 + 4000⁠. For C(18)C(18)⁠: 0.007×5832=40.8240.007\times 5832 = 40.824 and 0.003×324=0.9720.003\times 324 = 0.972⁠.)

Common mistakes
  1. Answering with the total cost. C(17)=4288.524C(17) = 4288.524 is the cost of 17 units, not the marginal cost.
  2. Decimal slips. 0.003×2=0.0060.003\times 2 = 0.006⁠, not 0.06; writing 0.06 would give 6.069−1.02+15=20.0496.069 - 1.02 + 15 = 20.049⁠.
  3. Keeping the fixed cost. The constant 4000 differentiates to 0.
  4. Forgetting the unit. The marginal cost is in rupees per unit of output.
What to take away
  1. Marginal cost = C′(x)C'(x)⁠: differentiate the total cost, then substitute the output level.
  2. Fixed costs disappear on differentiating; only the variable part of the cost affects the marginal cost.
  3. C′(x)C'(x) is close to the cost of one more unit, C(x+1)−C(x)C(x+1) - C(x)⁠, but not equal to it.

Every fact used, and where it comes from

FactUsed inSource
Marginal cost is the instantaneous rate of change of total cost with respect to outputStep 1Example 5, Section 6.2 (Lesson: rate of change of quantities)
ddx(xn)=nxn−1\frac{d}{dx}\left(x^n\right) = nx^{n-1}Step 2Table 5.3, Section 5.3
(u±v)′=u′±v′(u \pm v)' = u' \pm v'⁠; constants differentiate to 0Step 2Section 5.3, rule (1); Class XI, Limits and Derivatives
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