Internet Service Provider: Profit Maximum, Elasticity and a Per-Connection Tax
A Madurai internet service provider models its market with q measured in hundreds of connections, price p in rupees per connection per month, and revenue and cost in hundreds of rupees. Market research gives p = 200 - q, so revenue is R(q) = 200q - q^2. Monthly operating cost is C(q) = 3000 + 40q + 0.25q^2, where the constant is the fixed cost of the licensed head-end. The promoter currently runs at q = 50, where profit is still rising, and wants a defensible optimum rather than a rule of thumb. She also wants to know how sensitive demand is at the optimum, because a rival has just advertised a cheaper package. Separately, the state government is about to levy a tax of Rs 10 per connection per month, collected from the operator, and she wants the effect on output, price and profit before the announcement is made.
1. Find the output that maximises monthly profit, the corresponding price and the maximum profit, and verify using the second derivative that it is a maximum.
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Profit = R - C = (200q - q^2) - (3000 + 40q + 0.25q^2) = -1.25q^2 + 160q - 3000.
First derivative: dP/dq = -2.5q + 160. Setting it to zero gives q = 64 (hundred connections).
Second derivative: d2P/dq2 = -2.5 < 0, so q = 64 is a maximum, not a minimum or an inflection point.
Price: p = 200 - 64 = Rs 136 per connection per month.
Profit: R = 136 x 64 = 8704 hundred rupees; C = 3000 + 40(64) + 0.25(4096) = 3000 + 2560 + 1024 = 6584 hundred rupees; profit = 8704 - 6584 = 2120 hundred rupees, i.e. Rs 2,12,000 per month.
Check by the marginal rule: MR = 200 - 2q = 72 and MC = 40 + 0.5q = 72 at q = 64, consistent. Operating at q = 50 leaves money on the table: MR there is 100 against MC of 65, so each extra connection adds about Rs 35 of profit.
2. Compute the price elasticity of demand at the profit-maximising output and explain what it implies for a rival-induced price cut.
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From p = 200 - q we get dq/dp = -1, so elasticity = -(dq/dp)(p/q) = p/q = 136/64 = 2.125.
Demand is elastic at the optimum (2.125 > 1): a 1 per cent price cut raises quantity by about 2.1 per cent, so total revenue rises. That does not mean the promoter should cut price, because profit depends on cost as well. The marginal revenue condition makes the point precisely: MR = p(1 - 1/e) = 136(1 - 1/2.125) = 136 x 0.5294 = Rs 72, which equals MC. A price cut would raise revenue but also raise cost faster at the margin, reducing profit.
The practical reading for the rival threat: because demand is elastic, a small undercut by the rival can take a large share of connections, so the response should be service quality or a bundled offer rather than matching the price cut rupee for rupee. If she must cut price, she should re-solve the profit condition rather than assume that a lower price means a better outcome.
3. Work out the effect of the proposed Rs 10 per connection tax on the optimal output, price, profit and tax revenue, and state how much of the tax is passed to subscribers.
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The tax is Rs 10 per connection; since q is in hundreds, the tax bill is 10 x 100q rupees = 10q hundred rupees, so cost becomes 3000 + 40q + 0.25q^2 + 10q.
New marginal cost: 50 + 0.5q. Equate to MR: 200 - 2q = 50 + 0.5q, so 2.5q = 150 and q = 60.
New price: p = 200 - 60 = Rs 140. New profit: R = 140 x 60 = 8400; total cost = 3000 + 2400 + 900 + 600 = 6900; profit = 1500 hundred rupees = Rs 1,50,000.
Tax revenue = 10 x 60 = 600 hundred rupees = Rs 60,000 per month.
Effects: output falls by 4 hundred connections (400 subscribers), price rises by Rs 4 on a Rs 10 tax, so 40 per cent of the tax is passed on and 60 per cent is absorbed by the operator through a profit fall of Rs 62,000 (from 2,120 to 1,500 hundred rupees). Because the profit loss slightly exceeds the revenue collected, the tax also destroys some surplus; the promoter's best preparation is to model a Rs 10 rise as roughly a Rs 4 retail increase and a 6 per cent loss of subscribers.