- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
Functions of two variables and partial derivatives practice question 10
A company makes two models of e-bike, which compete for the same buyers. To sell of the first model and of the second in a month, it must charge dollars for the first and dollars for the second. Its monthly revenue is dollars. Find the marginal revenue from the first model, .
Revenue comes from both models. Does the second model’s revenue change when changes? Look at what its price depends on.
Multiply out first: . Each model’s revenue contributes an term.
Hold fixed and differentiate in : and are constants, and is times .
A formula in x and y. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §15.1Key results
Worked solution
Strategy
Write the revenue as a polynomial in x and y, then differentiate in x with y held fixed. The second model’s revenue depends on x too, because its price does.
Working
Revenue is quantity times price, for each model.
Each model contributes −xy, because each price falls when more of the other model is sold.
With y fixed, 1600y and −3y² are constants, and −2xy has x-derivative −2y.
At 100 and 50 bikes, one more of the first model adds about $1100 to revenue, less than its price there of $1550.
Answer
The answer is the partial derivative of R with respect to x equals 2000 minus 8x minus 2y, in dollars per bike.
Check
At (100, 50), one more of the first model raises revenue by R(101, 50) − R(100, 50) = $1096, close to 2000 − 800 − 100 = 1100.
Where this goes wrong
Differentiating only x p₁ and getting 2000 − 8x − y. The second model’s price falls as x rises, so its revenue y p₂ changes with x too, and that adds another −y.
Leans on
The power rule, from Differentiation rules
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