- Topic
- Optimization
- Difficulty
- Challenging
- Suggested time
- 7 min
- Marks
- 4
Optimization practice question 15
A food truck sells 400 burritos a week at $11 each, and estimates that every $1 cut in price would sell 40 more a week. Each burrito costs $3 to make, and the truck has fixed costs of $1200 a week. Assuming demand is linear in price, find the weekly sales and the price that maximise weekly profit.
Enter the price, in dollars.
Sales change linearly with the price. Which facts in the story give a point on that line and its slope, and what comes off the revenue to leave profit?
Write sales as a linear function of price. The line passes through , and its slope is : each dollar off the price adds 40 sales.
Profit is revenue minus costs: . Differentiate with the product rule.
A number.
Theory UBC Math 100 §8CLP-1 §3.5Key results
Worked solution
Strategy
Fit the demand line to the data, write profit as a function of price, differentiate, and justify the maximum. The fixed cost shifts the profit but not where it peaks.
Working
Slope −40 through (11, 400); sales cannot be negative, so the price is at most $21.
Each burrito earns p − 3 over its cost, and the fixed costs come off the total.
Product rule: both factors depend on p.
P is concave down everywhere, so p = 12 is the global maximum.
Sell 360 a week at $12, for a weekly profit of $2040.
Answer
The answer is charge 12 dollars and sell 360 burritos a week, for a weekly profit of 2040 dollars.
Check
Profit is 8 · 400 − 1200 = $2000 at $11 and 10 · 320 − 1200 = $2000 at $13, both below the $2040 at $12.
Where this goes wrong
Maximising revenue instead of profit: p(840 − 40p) peaks at $10.50, but every burrito sold also costs $3. Subtract the costs before differentiating; the unit cost moves the best price, the fixed cost does not.
Leans on
The equation of a line from a point and a slope, from Functions, graphs and inverses
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