Printing Club Pricing and Capacity Constraint
A youth printing club plans to produce revision booklets for a term. A market survey suggests that if the club charges p kwacha per booklet, monthly demand will be q = 1000 - 20p booklets, valid for 20 ≤ p ≤ 50. The club pays a fixed monthly overhead of K1,120 and a variable cost of K20 per booklet (paper, ink and binding). At a committee meeting, three proposals are debated: charge the lowest price that avoids a loss, charge the price that maximises profit, or charge a price that raises at least K3,000 profit while keeping the booklet below K40. Members disagree because the survey equation gives two break-even prices, and one of them looks unrealistic. The treasurer also warns that the model assumes demand falls steadily and that costs stay constant. The club has enough capacity for 550 booklets per month, not 700 as first assumed.
1. Derive the monthly profit function P(p) and solve P(p)=0. Which break-even price is actually feasible if the club can produce at most 550 booklets per month? Explain.
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Revenue is p(1000-20p). Variable cost is 20(1000-20p), and fixed cost is 1120. So P(p)=p(1000-20p)-20(1000-20p)-1120=(p-20)(1000-20p)-1120=-20p^2+1400p-21120. Set P=0: -20p^2+1400p-21120=0, so p^2-70p+1056=0, which factorises as (p-22)(p-48)=0. Thus p=22 or p=48. At p=22, demand is q=1000-440=560, which exceeds capacity 550, so that break-even cannot be served. At p=48, demand is q=40, which is feasible. The feasible break-even price is therefore K48; the lower root is a mathematical solution but not an operational one.
2. Find the price that maximises profit and the maximum monthly profit. Confirm that the capacity constraint is satisfied at that price.
Read the worked answer
P(p)=-20p^2+1400p-21120. The vertex is at p=-1400/(2×-20)=35, or complete the square: P=-20(p-35)^2+3380. Maximum profit is K3,380 at p=35. At that price demand is q=1000-700=300, which is below 550. The vertex is feasible. The endpoint p=50 gives zero demand and a loss, and p=20 gives demand 600, which violates capacity, so the interior maximum is the relevant one.
3. The committee wants a profit of at least K3,000 while keeping the price below K40. Solve the inequality and state the price range, taking the capacity limit into account.
Read the worked answer
P(p)≥3000 gives -20p^2+1400p-21120≥3000, so -20p^2+1400p-24120≥0. Divide by -20 and reverse the inequality: p^2-70p+1206≤0. Roots are (70±sqrt(4900-4824))/2=(70±sqrt76)/2≈(70±8.718)/2, giving p≈30.64 or 39.36. Thus 30.64≤p≤39.36. With p<40, the range is about [30.64,39.36). Capacity requires p≥22.5, which is already satisfied. Any price from about K30.64 to K39.36 meets the profit target, the price ceiling and capacity.