Pricing snack packs under capacity and a per-pack fee
A school canteen is introducing a new snack pack. A trial in a similar school suggests that at a price of Rs p per pack, daily demand is x = 240 - 2p packs, so the inverse demand relation is p = 120 - 0.5x. The canteen's variable cost is Rs 20 per pack and it incurs a fixed daily cost of Rs 200 for staff and utilities. The canteen currently has shelf and staff capacity for at most 80 packs per day. The manager wants to choose a price that maximises daily profit, but is unsure whether to use the unconstrained calculus solution or the capacity limit. The school council is also considering a municipal fee of Rs 10 per pack sold, which would be paid by the canteen.
1. Write the daily profit as a function of x, find the unconstrained profit-maximising output and price, and verify that it is a maximum.
Read the worked answer
Revenue = x times p = x(120 - 0.5x) = 120x - 0.5x^2.
Cost = 200 + 20x.
Profit P(x) = 120x - 0.5x^2 - (200 + 20x) = -0.5x^2 + 100x - 200.
Differentiate: P'(x) = -x + 100. Setting P'(x) = 0 gives x = 100.
P''(x) = -1, which is negative, so x = 100 gives a maximum.
The price is p = 120 - 0.5(100) = 70. The maximum profit is P(100) = -0.5(10,000) + 10,000 - 200 = Rs 4,800.
2. Now impose the capacity limit of 80 packs per day. Find the best feasible output, price and profit, and explain why the calculus condition P'(x) = 0 is not applied directly.
Read the worked answer
The unconstrained optimum is x = 100, which lies outside the feasible interval x <= 80. On the interval up to 80, P'(x) = 100 - x is positive, so profit is increasing as x rises toward 80. The best feasible choice is therefore the boundary x = 80.
Price: p = 120 - 0.5(80) = 80.
Profit: P(80) = -0.5(6,400) + 100(80) - 200 = -3,200 + 8,000 - 200 = Rs 4,600.
The stationary point x = 100 cannot be used because it is infeasible. At x = 80 the marginal profit is 100 - 80 = Rs 20 per pack, so the canteen would gain by selling more, but capacity prevents it.
3. Suppose the municipal fee of Rs 10 per pack is imposed. Find the new unconstrained optimum and the best feasible choice under the 80-pack capacity. Compare with the no-fee case.
Read the worked answer
The fee adds Rs 10 to the variable cost, so cost becomes 200 + 30x.
New profit = 120x - 0.5x^2 - 200 - 30x = -0.5x^2 + 90x - 200.
New derivative: -x + 90. Setting it to zero gives x = 90, with price p = 120 - 0.5(90) = 75 and unconstrained profit = -0.5(8,100) + 90(90) - 200 = Rs 3,850.
The capacity limit still binds because 90 > 80. The best feasible output is x = 80, price Rs 80, and profit = -0.5(6,400) + 90(80) - 200 = -3,200 + 7,200 - 200 = Rs 3,800.
Compared with the no-fee case, profit at the best feasible output falls from Rs 4,600 to Rs 3,800. The fee lowers the marginal profit but does not remove the capacity constraint; at x = 80 the marginal profit is still 90 - 80 = Rs 10 per pack.