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Chapter 6 · Application of Derivatives

Example 6: Marginal revenue of 3x2+36x+53x^2+36x+5 at x=5x = 5

ExamplesRate 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)=3x2+36x+5\mathrm{R}(x) = 3x^2 + 36x + 5⁠. Find the marginal revenue, when x=5x = 5⁠, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.

Exam answer

Marginal revenue MR=dRdx\mathrm{MR} = \frac{d\mathrm{R}}{dx}⁠. So

MR=6x+36\mathrm{MR} = 6x + 36

When x=5x = 5⁠,

MR=6(5)+36=66\mathrm{MR} = 6(5) + 36 = 66

Hence the required marginal revenue is ₹66.

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

The one idea this problem is built on

Marginal revenue is defined in the question as a rate of change with respect to the number of items sold, so it is the derivative R′(x)\mathrm{R}'(x)⁠. Differentiate the revenue function and evaluate at x=5x = 5⁠.

Given, to find, and the plan
Given
R(x)=3x2+36x+5\mathrm{R}(x) = 3x^2 + 36x + 5 rupees from selling xx units.
To find
The marginal revenue MR=dRdx\mathrm{MR} = \dfrac{d\mathrm{R}}{dx} at x=5x = 5⁠.
Plan
Differentiate term by term; substitute x=5x = 5⁠; interpret in rupees per unit; compare with actual revenues.

Basics you need

1A rate of change is a derivative

The rate of change of R\mathrm{R} with respect to xx is dRdx\frac{d\mathrm{R}}{dx} (Section 6.2), measured in rupees per unit sold. Basics: rate of change as a derivative

2Differentiating a polynomial

ddx(x2)=2x\frac{d}{dx}(x^2) = 2x⁠, ddx(x)=1\frac{d}{dx}(x) = 1⁠, the derivative of a constant is 0, and constant multipliers stay in front (Table 5.3 and the algebra of derivatives, Section 5.3).

Solution

  1. Differentiate the revenue function

    MR=dRdx=3(2x)+36(1)+0=6x+36.\mathrm{MR} = \frac{d\mathrm{R}}{dx} = 3(2x) + 36(1) + 0 = 6x + 36 .
    The question defines marginal revenue as the rate of change of total revenue with respect to the number sold, which is the derivative. The constant 5 does not change with xx⁠, so it adds nothing to the rate.
    Power rule and constant multiple rule on each term. Basics: rate of change as a derivative

  2. Substitute x=5x = 5

    MR(5)=6(5)+36=30+36=66.\mathrm{MR}(5) = 6(5) + 36 = 30 + 36 = 66 .

    So the marginal revenue at x=5x = 5 is ₹66 per unit.

    Units: R\mathrm{R} in rupees divided by xx in units gives rupees per unit. The value is positive, so revenue is still rising with sales at this level.
    Lesson: rate of change of quantities

  3. What the number means

    The actual revenues are

    R(4)=197,R(5)=260,R(6)=329.\mathrm{R}(4) = 197, \quad \mathrm{R}(5) = 260, \quad \mathrm{R}(6) = 329 .

    So the 5th unit brings in 260−197=63260 - 197 = 63 rupees and the 6th unit brings in 329−260=69329 - 260 = 69 rupees. The marginal revenue 66 at x=5x = 5 lies between, and is exactly their average:

    R(6)−R(4)2=329−1972=66.\frac{\mathrm{R}(6) - \mathrm{R}(4)}{2} = \frac{329 - 197}{2} = 66 .
    For a quadratic, the balanced quotient R(5+h)−R(5−h)2h\frac{\mathrm{R}(5+h) - \mathrm{R}(5-h)}{2h} equals R′(5)\mathrm{R}'(5) for every hh⁠: the h2h^2 terms cancel. That is why the average of 63 and 69 is exactly 66, not just close to it.
    R(5+h)−R(5)=66h+3h2\mathrm{R}(5+h) - \mathrm{R}(5) = 66h + 3h^2⁠; divide by hh and let h→0h \to 0 to get 66, the derivative.
Result

MR=R′(x)=6x+36\mathrm{MR} = \mathrm{R}'(x) = 6x + 36⁠, so when x=5x = 5

MR(5)=66 rupees per unit.\mathrm{MR}(5) = 66\ \text{rupees per unit}.

This agrees with the book's solution.

Check it with numbers

R(5)=3(25)+36(5)+5\mathrm{R}(5) = 3(25) + 36(5) + 5 =75+180+5=260= 75 + 180 + 5 = 260⁠. With a small step h=0.001h = 0.001⁠:

R(5.001)−R(5)0.001=66+3(0.001)=66.003,\begin{aligned} \frac{\mathrm{R}(5.001) - \mathrm{R}(5)}{0.001} &= 66 + 3(0.001) \\ &= 66.003 , \end{aligned}

which tends to 66 as the step shrinks.

Common mistakes
  1. Giving R(5)=260\mathrm{R}(5) = 260, the total revenue, instead of its rate of change.
  2. Differentiating 3x23x^2 as 3x3x or 6x26x^2⁠. The power rule gives 3⋅2x=6x3 \cdot 2x = 6x⁠.
  3. Treating marginal revenue as average revenue: 2605=52\frac{260}{5} = 52 rupees per unit is the average, a different quantity.
What to take away
  1. Marginal revenue = dRdx\frac{d\mathrm{R}}{dx}⁠, in rupees per unit.
  2. It sits between the revenue from the last unit and the next one.
  3. Marginal cost (Example 5) and marginal revenue are the same idea: a derivative with respect to the number of units.

Every fact used, and where it comes from

FactUsed inSource
Marginal revenue is the rate of change of revenue with respect to units soldStep 1Stated in the question; Section 6.2 (Lesson: rate of change of quantities)
Power, constant multiple and sum rulesStep 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 197, 260, 329, 63, 69, 66.003Step 3; CheckArithmetic, recomputed in the site's verification script
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