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Chapter 6 · Application of Derivatives

Exercise 6.1 · Q18 (MCQ): Marginal revenue of 3x2+36x+53x^2 + 36x + 5 at x=15x = 15

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

Choose the correct answer.

The total revenue in Rupees received from the sale of xx units of a product is given by

R(x)=3x2+36x+5.R(x) = 3x^2 + 36x + 5.

The marginal revenue, when x=15x = 15 is

(A) 116    (B) 96    (C) 90    (D) 126

Exam answer

R′(x)=6x+36R'(x) = 6x + 36⁠.

R′(15)=6(15)+36=126R'(15) = 6(15) + 36 = 126⁠.

Answer: (D) 126126

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

The one idea this problem is built on

Marginal revenue is the rate of change of total revenue with respect to the number of units sold, that is, the derivative R′(x)R'(x) (Example 6, Section 6.2, uses this very same RR at x=5x = 5⁠). Here R′(x)=6x+36R'(x) = 6x + 36⁠, so at x=15x = 15 it is 90+36=12690 + 36 = 126⁠, option (D).

Given, to find, and the plan
Given
R(x)=3x2+36x+5R(x) = 3x^2 + 36x + 5 rupees from selling xx units.
To find
The marginal revenue R′(15)R'(15)⁠, and which option it matches.
Plan
(1) Marginal revenue means R′(x)R'(x)⁠. (2) Differentiate. (3) Substitute x=15x = 15⁠. (4) Examine every option.

Basics you need

1Marginal revenue is a derivative

MR =dRdx= \frac{dR}{dx}⁠, the rate of change of total revenue with respect to the number of items sold (Example 6, Section 6.2). Basics: rate of change as a derivative

2Differentiating a polynomial

Term by term: ddx(3x2)=6x\frac{d}{dx}\left(3x^2\right) = 6x⁠, ddx(36x)=36\frac{d}{dx}(36x) = 36⁠, ddx(5)=0\frac{d}{dx}(5) = 0⁠.

Solution

  1. Differentiate the revenue

    MR=R′(x)=3(2x)+36+0=6x+36.\begin{aligned} \text{MR} = R'(x) &= 3(2x) + 36 + 0 \\ &= 6x + 36. \end{aligned}
    Marginal revenue is defined as the rate of change of revenue with respect to units sold, so it is the derivative of RR⁠. Each term is differentiated by the power rule; the constant 5 contributes nothing. This matches the book's working in Example 6, where dRdx=6x+36\frac{dR}{dx} = 6x + 36⁠.
    ddx(xn)=nxn−1\frac{d}{dx}\left(x^n\right) = nx^{n-1} (Table 5.3) and the sum rule (Section 5.3, rule (1)). Basics: rate of change as a derivative

  2. Substitute x=15x = 15

    R′(15)=6(15)+36=90+36=126.\begin{aligned} R'(15) &= 6(15) + 36 \\ &= 90 + 36 = 126. \end{aligned}

    So the marginal revenue is ₹126 per unit when 15 units are sold.

    Differentiate first, then substitute. Putting x=15x = 15 into RR itself would give the total revenue R(15)=675+540+5=1220R(15) = 675 + 540 + 5 = 1220⁠, which answers a different question (and is not among the options).

  3. Examine every option

    OptionValueEqual to R′(15)=126R'(15) = 126⁠?Why it fails or holds
    (A)116No6x+36=1166x + 36 = 116 needs x=403x = \frac{40}{3}⁠, not 15; it is not the MR at any whole number of units
    (B)96No6(10)+36=966(10) + 36 = 96⁠: the MR at x=10x = 10⁠, not 15
    (C)90No6(15)=906(15) = 90 only: the 36 (from the 36x36x term) has been dropped
    (D)126Yes6(15)+36=1266(15) + 36 = 126

    Only (D) is correct.

    Since R′(x)=6x+36R'(x) = 6x + 36 takes each value at exactly one xx⁠, any number other than 126 is the marginal revenue at some other sales level (or at none), never at x=15x = 15⁠. Option (C) is the tempting slip of forgetting that 36x36x differentiates to 36.
    Marginal revenue at a point: evaluate R′R' there. Lesson: rate of change of quantities
Result

R′(15)=6(15)+36=126R'(15) = 6(15) + 36 = 126⁠, so the marginal revenue is ₹126. The answer is (D) 126. (Textbook answer: D.)

Check it with numbers

R(14)=3(196)+36(14)+5=1097R(14) = 3(196) + 36(14) + 5 = 1097 and R(16)=3(256)+36(16)+5=1349R(16) = 3(256) + 36(16) + 5 = 1349⁠. For a quadratic, the symmetric difference gives the derivative at the midpoint exactly:

R(16)−R(14)2=1349−10972=2522=126.\frac{R(16) - R(14)}{2} = \frac{1349 - 1097}{2} = \frac{252}{2} = 126.
Common mistakes
  1. Dropping the constant from 36x36x⁠. ddx(36x)=36\frac{d}{dx}(36x) = 36⁠, not 0; dropping it gives option (C), 90.
  2. Computing R(15)R(15)⁠. 12201220 is the total revenue, not the marginal revenue.
  3. Substituting the wrong xx⁠. 96 is the marginal revenue at 10 units.
What to take away
  1. Marginal revenue = R′(x)R'(x)⁠; differentiate first, then substitute.
  2. A linear term bxbx differentiates to the constant bb⁠, and a constant term to 0.
  3. For a quadratic revenue, R(a+1)−R(a−1)2\frac{R(a+1) - R(a-1)}{2} is a quick exact check of R′(a)R'(a)⁠.

Every fact used, and where it comes from

FactUsed inSource
Marginal revenue is the rate of change of total revenue with respect to the number of items soldSteps 1, 2Example 6, Section 6.2 (Lesson: rate of change of quantities)
ddx(xn)=nxn−1\frac{d}{dx}\left(x^n\right) = nx^{n-1}Step 1Table 5.3, Section 5.3
Sum rule; constants differentiate to 0Step 1Section 5.3, rule (1); Class XI, Limits and Derivatives
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