- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Optimization in two variables and Lagrange multipliers practice question 3
A student’s utility from coffees and sandwiches a month is . A coffee costs $2, a sandwich costs $5, and the monthly budget is $100. What is the largest utility the budget allows?
What is being maximised, and what is the constraint? Turn the two prices and the $100 budget into an equation in and .
The budget is . The Lagrange conditions and , with , read and .
Substitute and into the budget to find , then and , then .
A whole number.
Theory UBC Math 100 §16.3Key results
Worked solution
Strategy
Maximise U subject to the budget 2x + 5y = 100. Lagrange multipliers: each partial derivative of U is λ times the matching partial derivative of the budget expression, and the budget itself must hold.
Working
The three Lagrange equations, with g(x, y) = 2x + 5y.
Substitute both into the budget.
That is $50 on each good: with U = xy the budget is split evenly.
At either end of the budget line one good is not bought and U = 0, so this single candidate is the maximum.
Answer
The answer is 250, with 25 coffees and 10 sandwiches.
Check
Other bundles that spend the whole $100, such as 20 coffees with 12 sandwiches or 30 with 8, give only 240.
Where this goes wrong
Stopping at x = 25 or at λ = 5. The question asks for the utility itself, which is U(25, 10) = 250.
Leans on
Optimising with a constraint, from Optimization
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