Tea Stall Profit, Revenue and a Demand Shift
A roadside tea stall in Jorhat sells x cups of tea per day at price p rupees per cup, where p=40−0.05x for 0≤x≤600. The daily cost is C(x)=200+8x+0.01x^2. The owner currently sells 300 cups at ₹25 each and argues that raising the price to ₹30 must increase revenue because each cup earns more. The stall can serve at most 500 cups per day because of staff and water constraints. A new college canteen is expected to reduce demand at every price by 20 cups, so the owner models the new demand as p=40−0.05(x+20). He wants to know whether 300 cups is profit-maximising, what output would maximise revenue rather than profit, and how the new canteen changes the profit-maximising output if the capacity constraint remains 500 cups. He is unsure whether the endpoints of the interval or the derivative condition should decide the answer.
1. Find the profit function and the profit-maximising daily output under the 500-cup capacity. Is 300 cups optimal? Compare profits.
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Profit is π(x)=xp−C(x)=x(40−0.05x)−(200+8x+0.01x^2). Expanding: π(x)=40x−0.05x^2−200−8x−0.01x^2 = −200+32x−0.06x^2. Differentiate: π'(x)=32−0.12x, so the critical point is x=32/0.12=266.67. Since π''(x)=−0.12<0, this is a maximum. The domain is [0,500]; the endpoints give π(0)=−200 and π(500)=−200+16000−15000=800, both below π(266.67)=−200+8533.33−4266.67=4066.67. Thus 300 cups is not optimal. π(300)=−200+9600−5400=4000, so the loss from producing 300 instead of about 267 is roughly ₹66.67 per day. The capacity of 500 does not bind.
2. Find the revenue-maximising output and price, and explain why it differs from the profit-maximising output.
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Revenue is R(x)=xp=40x−0.05x^2. Then R'(x)=40−0.1x, giving x=400, and R''(x)=−0.1<0, so this is a maximum. The corresponding price is p=40−0.05(400)=₹20. Revenue at 400 cups is 400×20=₹8000, while at 300 cups it is 300×25=₹7500. Profit is maximised at about 267 cups because costs rise with output; revenue ignores costs and keeps rising until marginal revenue is zero. Hence the revenue-maximising output is larger than the profit-maximising output.
3. With the new demand p=40−0.05(x+20), find the new profit-maximising output and compare it with the old one. Does the 500-cup capacity bind?
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The new price simplifies to p=40−0.05x−1=39−0.05x. New profit is π(x)=x(39−0.05x)−(200+8x+0.01x^2)=−200+31x−0.06x^2. Then π'(x)=31−0.12x, so x=31/0.12=258.33. Since π''(x)=−0.12<0, this is a maximum. It lies inside [0,500], so the capacity does not bind. The optimal output falls from about 267 cups to about 258 cups, a decrease of about 8–9 cups, because the demand curve has shifted down by ₹1 at every quantity. The new price at the optimum is about ₹26.08.