Case Study 2 — Pricing a Streaming Plan: Margins and Surplus

DataField.Dev

Case Study 2 — Pricing a Streaming Plan: Margins and Surplus

Field: Economics (Modeling Portfolio Track B) Calculus used: The tangent problem as marginal revenue (§1.1, §1.7); the area problem as consumer surplus (§1.2) Key skills demonstrated: Reading a derivative as a "marginal" quantity; reading an integral as an accumulated economic total; seeing why the word marginal literally means the derivative

The Setup

You run pricing for a new streaming service. Market research gives you a demand curve: a relationship between the monthly price $p$ you charge and the number of subscribers $q$ you can expect to sign up. After fitting the survey data, your analysts hand you

$$p(q) = 12 - 0.002\,q \qquad \text{(dollars per month).}$$

Read it as a willingness-to-pay schedule. The very first, most enthusiastic subscriber would pay nearly \$12/month. To attract the 2{,}000th subscriber you have to drop the price to $p(2000) = 12 - 0.002(2000) = \$8$. The more subscribers you want, the lower the price you must set — a downward-sloping demand curve, the most basic object in microeconomics.

Two questions land on your desk. If we already have 1{,}000 subscribers, how much extra revenue does signing up one more bring in? And: at a price of \$8, how much value are subscribers getting that they did not have to pay for? The first is a tangent problem in disguise; the second is an area problem. Both are calculus, and both yield to the two ideas from this chapter.

The Tangent Problem: Marginal Revenue

Your monthly revenue is price times quantity. Since the price you can charge depends on how many subscribers you want, revenue is a function of $q$ alone:

$$R(q) = p(q)\cdot q = (12 - 0.002\,q)\,q = 12q - 0.002\,q^2.$$

Economists obsess over marginal revenue: the extra revenue from one more unit sold. Naively you might compute $R(1001) - R(1000)$ — the revenue change from exactly one more subscriber. But the clean, theory-grade definition is the instantaneous rate of change of revenue with respect to quantity — the slope of the revenue curve at $q = 1000$. That is the tangent problem from §1.1, with $q$ playing the role of $x$.

Form the secant slope and shrink the step $\Delta q$:

$$\text{MR} \approx \frac{R(1000 + \Delta q) - R(1000)}{\Delta q}.$$

$\Delta q$	secant slope (\$/subscriber)
$10$	$7.980$
$1$	$7.998$
$0.1$	$7.9998$
$0.01$	$7.99998$

The slopes converge to \$8 per additional subscriber. We can nail the limit exactly with the §1.7 algebra:

$$\frac{R(q+\Delta q) - R(q)}{\Delta q} = \frac{\big[12(q+\Delta q) - 0.002(q+\Delta q)^2\big] - \big[12q - 0.002q^2\big]}{\Delta q}.$$

Expanding $(q+\Delta q)^2 = q^2 + 2q\,\Delta q + \Delta q^2$ and cancelling, the numerator becomes $12\,\Delta q - 0.002(2q\,\Delta q + \Delta q^2)$, so the quotient simplifies to

$$12 - 0.004q - 0.002\,\Delta q \;\xrightarrow{\;\Delta q \to 0\;}\; 12 - 0.004q.$$

This is the marginal-revenue function, $\text{MR}(q) = 12 - 0.004q$. At $q = 1000$ it gives $12 - 4 = \$8$, matching the table. Notice it falls twice as steeply as the demand curve — a standard and important fact every economics student eventually proves, and here it dropped out of a single limit.

# Marginal revenue as the slope of R(q) = 12q - 0.002 q^2 at q = 1000.
def R(q: float) -> float:
    return 12 * q - 0.002 * q**2

q0 = 1000.0
for dq in [10, 1, 0.1, 0.01]:
    mr = (R(q0 + dq) - R(q0)) / dq
    print(f"dq = {dq:<6} marginal revenue = ${mr:.5f}")

# Output:
# dq = 10     marginal revenue = $7.98000
# dq = 1      marginal revenue = $7.99800
# dq = 0.1    marginal revenue = $7.99980
# dq = 0.01   marginal revenue = $7.99998

Why "marginal" means "derivative." In economics, marginal cost, marginal revenue, and marginal utility are all derivatives — the instantaneous rate of change of cost, revenue, or satisfaction as you produce or consume one more unit. The whole vocabulary of marginal analysis is calculus in disguise, and the secant-slope limit you just computed is the engine underneath it. Chapter 10 uses exactly this idea to find the profit-maximizing price: set marginal revenue equal to marginal cost.

The Area Problem: Consumer Surplus

Now the second question. Suppose you set the price at $p^* = \$8$. From the demand curve, $8 = 12 - 0.002q$ gives $q^ = 2000$ subscribers. Everyone pays the same \$8. But the demand curve says many of those subscribers would gladly have paid more — the first subscriber valued the service near \$12, yet pays only \$8 and pockets the \$4 difference. That pocketed value, summed over all subscribers, is the consumer surplus*: the total benefit subscribers receive beyond what they pay.

Geometrically, consumer surplus is the area between the demand curve and the horizontal price line, from $q = 0$ to $q = q^*$. That is precisely the area problem of §1.2 — the area of a region with a slanted top edge. We attack it with Riemann rectangles: slice $[0, 2000]$ into $n$ strips, and on each strip the "extra value per subscriber" is $p(q) - 8 = (12 - 0.002q) - 8 = 4 - 0.002q$. Multiply by the strip width (a small batch of subscribers) and sum:

$$\text{CS} \approx \sum_{i=1}^{n} \big[\,4 - 0.002\,q_i\,\big]\cdot \frac{2000}{n}.$$

# Consumer surplus = area between demand and the $8 price line, q in [0, 2000].
def extra_value(q: float) -> float:
    return (12 - 0.002 * q) - 8        # willingness to pay above the price

def surplus_estimate(n: int) -> float:
    width = 2000.0 / n
    return sum(extra_value(i * width) * width for i in range(1, n + 1))

for n in [4, 10, 100, 1000]:
    print(f"n = {n:5d}   consumer surplus estimate = ${surplus_estimate(n):,.2f}")

# Output:
# n =     4   consumer surplus estimate = $3,000.00
# n =    10   consumer surplus estimate = $3,600.00
# n =   100   consumer surplus estimate = $3,960.00
# n =  1000   consumer surplus estimate = $3,996.00

The estimates climb toward \$4{,}000 per month** of consumer surplus. Here the region happens to be a triangle — base $2000$ subscribers, height \$4 of surplus for the very first one — so geometry confirms the limit instantly: $\tfrac12 \times 2000 \times 4 = \$4{,}000$. The Riemann sum and the triangle formula agree, exactly as the rectangle sums agreed with the exact value in §1.2. (For a curved demand schedule there would be no triangle to fall back on, and the integral would be the only way through — which is the whole point of building integration in Chapter 13.)

Two Problems, One Pricing Decision

Step back and notice that a single business decision needed both halves of calculus:

Marginal revenue — a slope, the tangent problem — tells you how revenue responds to the next subscriber. It is a local quantity: the behaviour of the revenue curve right at $q = 1000$.
Consumer surplus — an area, the area problem — tells you the total value accumulated across all subscribers. It is a global quantity: a sum over the whole interval $[0, 2000]$.

Local and global, slopes and areas — the same dichotomy you met in §1.3, now driving a pricing strategy. And just as in §1.4, the two are secretly linked: the surplus is an accumulation, and accumulation is the inverse of taking a rate. When you reach Chapter 14 you will be able to compute that \$4{,}000 in one line by antidifferentiating $4 - 0.002q$ and evaluating at the endpoints — no thousand-rectangle sum required.

Connections to Other Chapters

Marginal cost/revenue/profit as derivatives: Chapter 5 (the derivative as a rate) and Chapter 7 (rules that make $\text{MR} = 12 - 0.004q$ immediate).
Maximizing profit by setting marginal revenue equal to marginal cost: Chapter 9 and Chapter 10 (optimization).
Consumer and producer surplus as definite integrals: Chapter 13 and Chapter 18 (applications of integration).
The antiderivative shortcut for the surplus area: Chapter 14 (Fundamental Theorem of Calculus).

Discussion Questions

We found marginal revenue $\text{MR}(q) = 12 - 0.004q$ falls twice as fast as demand $p(q) = 12 - 0.002q$. Using a picture of the revenue curve $R(q) = 12q - 0.002q^2$, explain why marginal revenue eventually turns negative, and what a firm should never do once it does.
The consumer-surplus region here was a triangle, so we could check the integral with geometry. Invent a curved demand schedule (for example $p(q) = 12 - 0.000001\,q^2$) for which no elementary area formula applies. Why does this force you to use integration rather than triangle-and-rectangle geometry?
Marginal revenue is a local measure and consumer surplus is a global measure. Map this onto the local/global contrast of §1.3. Which other pairs of "rate" and "total" quantities can you name from everyday economics?
Right-endpoint rectangles gave surplus estimates that rose toward \$4{,}000 from *below* ($3000, 3600, 3960, 3996$). Sketch the rectangles under the falling line $4 - 0.002q$ and explain why right-endpoint sums undershoot here, whereas in Case Study 1 they overshot.
A real streaming firm also has costs (servers, licensing, support). Sketch how you would add a cost curve $C(q)$ and use marginal analysis to choose the subscriber count that maximizes profit. Which chapter gives you the tool to finish the job?

Your Turn — Mini-Project

Using $p(q) = 12 - 0.002q$ and $R(q) = 12q - 0.002q^2$:

Write marginal_revenue(q0, dq) and tabulate MR at $q = 500$, $1500$, and $2500$ by shrinking dq. At which quantity does marginal revenue hit zero? (That quantity maximizes revenue — a preview of Chapter 10.)
Write consumer_surplus(price, n) that finds $q^*$ from the demand curve and Riemann-sums the surplus. Compute the surplus at prices \$6, \$8, and \$10. How does lowering the price change the surplus?
Plot the demand curve, draw the horizontal price line at \$8, shade the consumer-surplus triangle, and mark the point $(1000, p(1000))$ where you measured marginal revenue. One figure now holds both a tangent-problem reading and an area-problem reading of the same curve.

Case Study 2 — Pricing a Streaming Plan: Margins and Surplus

The Setup

The Tangent Problem: Marginal Revenue

The Area Problem: Consumer Surplus

Two Problems, One Pricing Decision

Connections to Other Chapters

Discussion Questions

Your Turn — Mini-Project

Further Reading