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Chapter 6 · Application of Derivatives

Exercise 6.1 · Q16: Marginal revenue of 13x2+26x+1513x^2 + 26x + 15 at x=7x = 7

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

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

R(x)=13x2+26x+15.R(x) = 13x^2 + 26x + 15.

Find the marginal revenue when x=7x = 7⁠.

Exam answer

Marginal revenue =R′(x)= R'(x)⁠.

R′(x)=26x+26R'(x) = 26x + 26 R′(7)=26(7)+26=208R'(7) = 26(7) + 26 = 208

∴\therefore the marginal revenue is ₹208208⁠.

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 items sold (the book's definition in Example 6). So it is the derivative R′(x)=26x+26R'(x) = 26x + 26⁠, evaluated at x=7x = 7⁠. It is not the revenue R(7)R(7) itself.

Given, to find, and the plan
Given
R(x)=13x2+26x+15R(x) = 13x^2 + 26x + 15 rupees from selling xx units.
To find
The marginal revenue MR=R′(x)\text{MR} = R'(x) at x=7x = 7⁠.
Plan
(1) Recall what marginal revenue means. (2) Differentiate RR⁠. (3) Substitute x=7x = 7⁠. (4) State the answer with units and interpret it.

Basics you need

1Marginal revenue is a derivative

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

2Differentiating a polynomial

Term by term: ddx(ax2)=2ax\frac{d}{dx}\left(ax^2\right) = 2ax⁠, ddx(bx)=b\frac{d}{dx}(bx) = b⁠, and a constant gives 0.

Solution

  1. Marginal revenue means R′(x)R'(x)

    By the definition in Example 6,

    MR=dRdx=R′(x),\text{MR} = \frac{dR}{dx} = R'(x),

    measured in rupees per unit sold.

    "Marginal" means the rate at the current level of sales: how fast the revenue is growing as one more item is sold. That is a derivative, not a value of RR⁠.
    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

    R′(x)=13(2x)+26+0=26x+26=26(x+1).\begin{aligned} R'(x) &= 13(2x) + 26 + 0 \\ &= 26x + 26 = 26(x + 1). \end{aligned}
    Differentiate each term: 13x213x^2 gives 13⋅2x=26x13\cdot 2x = 26x⁠; 26x26x gives 26; the constant 15 gives 0. The factored form 26(x+1)26(x+1) makes the substitution in the next step quick.
    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)).

  3. Substitute x=7x = 7

    R′(7)=26(7)+26=182+26=208.\begin{aligned} R'(7) &= 26(7) + 26 = 182 + 26 \\ &= 208. \end{aligned}

    Equivalently 26(7+1)=26×8=20826(7 + 1) = 26\times 8 = 208⁠.

    Both forms must give the same number, which is a quick check on the arithmetic.

  4. State and interpret the answer

    The marginal revenue when x=7x = 7 is ₹208 per unit: at a sales level of 7 units, revenue is rising at ₹208 for each additional unit sold.

    Compare the actual revenue from whole units: R(6)=639R(6) = 639⁠, R(7)=834R(7) = 834⁠, R(8)=1055R(8) = 1055⁠. The 7th unit brought in 834−639=195834 - 639 = 195 rupees and the 8th brings in 1055−834=2211055 - 834 = 221 rupees; the marginal revenue 208 lies between them, and is exactly their average.

    For a quadratic, the average of the steps on either side, R(8)−R(6)2\frac{R(8) - R(6)}{2}⁠, equals the derivative at the middle point exactly (the x2x^2 terms cancel in a symmetric difference). That makes it a neat independent check here.
    Rate of change of revenue with respect to units sold. Lesson: rate of change of quantities
Result

Marginal revenue =R′(7)=26(7)+26== R'(7) = 26(7) + 26 = ₹208. (Textbook answer: ₹208.)

Check it with numbers

R(8)=13(64)+26(8)+15=1055R(8) = 13(64) + 26(8) + 15 = 1055 and R(6)=13(36)+26(6)+15=639R(6) = 13(36) + 26(6) + 15 = 639⁠, so

R(8)−R(6)8−6=1055−6392=4162=208,\frac{R(8) - R(6)}{8 - 6} = \frac{1055 - 639}{2} = \frac{416}{2} = 208,

which agrees with R′(7)=208R'(7) = 208⁠.

Common mistakes
  1. Answering with the revenue. R(7)=13(49)+26(7)+15R(7) = 13(49) + 26(7) + 15⁠, that is 637+182+15=834637 + 182 + 15 = 834⁠, is the total revenue, not the marginal revenue.
  2. Differentiating 13x213x^2 as 13x13x or 26x226x^2⁠. The power rule gives 2×13x=26x2\times 13x = 26x⁠.
  3. Keeping the constant 15. It differentiates to 0.
What to take away
  1. Marginal revenue = R′(x)R'(x)⁠: differentiate, then substitute the sales level.
  2. Marginal cost and marginal revenue are the same idea (a rate of change with respect to the number of units), applied to cost and to revenue.
  3. Comparing with R(x+1)−R(x)R(x+1) - R(x) is a useful sense check: the marginal revenue is close to the revenue from one more unit.

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 soldStep 1Example 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 2Table 5.3, Section 5.3
Sum rule; constants differentiate to 0Step 2Section 5.3, rule (1); Class XI, Limits and Derivatives
Symmetric difference of a quadratic equals its derivative at the midpointStep 4, CheckAlgebra (the x2x^2 terms cancel)
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