Bookstore Calculator Pricing and a Conditional Supplier Discount
A campus bookstore is planning its calculator order for the fall semester. The supplier charges 15 KD per calculator, and the bookstore pays a fixed 300 KD for shipping and display materials. From past semesters, the manager estimates that if the selling price is p KD, the number sold will be 200 - 5p, provided p is between 15 and 35 KD. The bookstore currently sells the calculator at 22 KD. The supplier has offered a 2 KD per-unit discount if the bookstore orders at least 80 calculators in one shipment. The discount would reduce the per-unit cost from 15 KD to 13 KD, but the fixed cost stays the same. The manager wants to know whether the discount is worth using and what price would maximize profit under each cost structure. Profit is revenue minus total cost, and unsold calculators have no salvage value. The bookstore also wants to avoid a loss on the order.
1. Ignoring the discount, express profit as a function of the selling price p. Find the profit-maximizing price and quantity, and compare with the current 22 KD price.
Read the worked answer
Regular unit cost is 15 KD, so profit is
P(p) = (p - 15)(200 - 5p) - 300.
Expanding gives P(p) = -5p^2 + 275p - 3300. This is a downward-opening quadratic, so its vertex is at p = 275/(2*5) = 27.5 KD. At that price, demand is 200 - 5(27.5) = 62.5 calculators, and continuous profit is 481.25 KD.
At the current price of 22 KD, demand is 90 and profit is (22 - 15)(90) - 300 = 330 KD. Thus 22 KD is not optimal; moving toward 27.5 KD raises profit by about 151 KD. Because calculators are discrete, check integer prices 27 and 28: each gives profit 480 KD, so either is a practical best integer price.
2. Now include the discount. Find the unconstrained profit-maximizing price and quantity at 13 KD cost. Check whether the 80-unit minimum is satisfied. Then find the best eligible discount plan and compare it with the regular-cost optimum.
Read the worked answer
With unit cost 13 KD, profit is
P(p) = (p - 13)(200 - 5p) - 300 = -5p^2 + 265p - 2900.
The vertex is at p = 265/(2*5) = 26.5 KD, giving demand 200 - 5(26.5) = 67.5 and profit 611.25 KD. But 67.5 is below 80, so this unconstrained plan does not qualify for the discount.
For an order q >= 80, the demand price is p = (200 - q)/5. Profit as a function of q is
P(q) = [((200 - q)/5) - 13]q - 300 = 27q - 0.2q^2 - 300.
This quadratic also peaks at q = 67.5, so for q >= 80 it decreases as q increases. Therefore the best eligible quantity is q = 80. The price is p = (200 - 80)/5 = 24 KD, and profit is (24 - 13)(80) - 300 = 580 KD. This beats the regular-cost optimum of 481.25 KD by 98.75 KD, so the discount is worth using, but the store should order 80 and price at 24 KD.
3. The manager says: 'The discount lowers unit cost, so we should take it and keep the selling price at 22 KD.' Evaluate that statement quantitatively.
Read the worked answer
At p = 22 KD with the discount, demand is 200 - 5(22) = 90 calculators. The unit margin is 22 - 13 = 9 KD, so profit is 9(90) - 300 = 510 KD. This is better than the current no-discount profit of 330 KD and better than the regular-cost optimum of 481.25 KD, so taking the discount does help.
However, 510 KD is 70 KD below the best eligible discount plan of 580 KD at p = 24 KD. Also, at p = 22 the store would want 90 calculators, which satisfies the 80-unit minimum. The statement is directionally right but not optimal: keeping the price at 22 KD leaves money on the table. The best discount plan is to order 80 calculators and sell at 24 KD.