Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, critical points
This covers three topics that appear in the free-response section of the MAT 211 final: Lagrange multipliers for constrained optimisation, implicit differentiation for equations that cannot be solved for y explicitly, and setting up objective functions for profit maximisation problems. Lagrange multipliers alone account for 10 points on this practice exam, so the method is worth learning step by step. Implicit differentiation and profit setup are smaller but reliably tested. You should already be comfortable with partial derivatives before tackling this material.
Lagrange multipliers let you optimise f(x, y) subject to a constraint g(x, y) = 0 by building a combined function L = f - λg and solving L_x = 0, L_y = 0, g = 0. Implicit differentiation finds dy/dx from an equation relating x and y without solving for y first. Profit maximisation starts by identifying the right objective function: Profit = Revenue - Cost = (price - unit cost) × quantity.
Lagrange multiplier (λ)
A new variable introduced to convert a constrained optimisation problem into an unconstrained system of equations. It represents the rate at which the optimal value of the objective function changes per unit change in the constraint.
Think of it as: the "price" of tightening or loosening the constraint by one unit.
Constraint
A condition that restricts the values of x and y. Written as g(x, y) = 0 (or equivalently, rearranged so one side is zero).
In simple terms, this means: you cannot choose x and y freely; they must satisfy a rule.
Objective function
The function you want to maximise or minimise. In profit problems, this is typically Profit = Revenue - Cost.
Think of it as: the thing you are trying to make as large (or small) as possible.
Implicit differentiation
A technique for finding dy/dx when x and y are tangled together in an equation and you cannot (or do not want to) isolate y. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
In simple terms, this means: differentiate everything in place, remember the chain rule for y terms, and then algebra your way to dy/dx.
Profit function
P = R - C, where R is total revenue and C is total cost. Revenue is price times quantity (R = pq). Cost may include fixed and variable components.
In simple terms, this means: what you take in minus what you spend.
Write the constraint in the form g(x, y) = 0.
Form the Lagrangian: L(x, y, λ) = f(x, y) - λ · g(x, y).
Take partial derivatives and set them to zero: L_x = 0, L_y = 0, and g(x, y) = 0.
Solve the resulting system of three equations for x, y, and λ.
Evaluate f at the solution to find the optimised value.
Maximise f(x, y) = xy subject to 3x + y = 12.
Step 1: Rewrite the constraint as g(x, y) = 3x + y - 12 = 0.
Step 2: L(x, y, λ) = xy - λ(3x + y - 12).
Step 3: Set partial derivatives to zero.
L_x = y - 3λ = 0 → y = 3λ
L_y = x - λ = 0 → x = λ
g(x, y) = 3x + y - 12 = 0
Step 4: Substitute x = λ and y = 3λ into the constraint.
3(λ) + 3λ - 12 = 0
6λ = 12
λ = 2
So x = 2 and y = 6.
Step 5: f(2, 6) = 2 × 6 = 12.
The maximum value of f(x, y) is 12, occurring at (2, 6).
Use implicit differentiation when x and y are mixed together in an equation and you want dy/dx without solving for y.
Differentiate every term with respect to x.
Whenever you differentiate a term involving y, multiply by dy/dx (chain rule).
Collect all dy/dx terms on one side.
Factor out dy/dx and solve.
Given 5xy + 2x = y², find dy/dx.
Step 1: Differentiate both sides with respect to x.
d/dx(5xy) + d/dx(2x) = d/dx(y²)
Step 2: Apply the product rule to 5xy and the chain rule to y².
5y + 5x(dy/dx) + 2 = 2y(dy/dx)
Step 3: Collect dy/dx terms on one side.
5x(dy/dx) - 2y(dy/dx) = -5y - 2
Step 4: Factor and solve.
dy/dx(5x - 2y) = -(5y + 2)
dy/dx = -(5y + 2) / (5x - 2y)
This exists provided 5x - 2y ≠ 0.
In a profit maximisation problem, the objective function is:
Profit = Revenue - Cost
Revenue = price × quantity = p × q.
Cost = (cost per unit) × quantity = (unit cost) × q.
So Profit = p × q - (unit cost) × q = (p - unit cost) × q.
Hercules Films sells q = 550,000 - 23,500p copies at price p. Each copy costs $5 to make.
Revenue: R = p × q
Cost: C = 5q
Profit = R - C = pq - 5q = (p - 5)q
The objective function is Profit = (p - 5)q.
Answer: B.
Note: to actually solve the optimisation, you would substitute q = 550,000 - 23,500p into the profit function, then take the derivative with respect to p and set it to zero. But the question only asks you to identify the correct objective function.
Lagrangian: L(x, y, λ) = f(x, y) - λ · g(x, y), where g(x, y) = 0 is the constraint.
Implicit differentiation of a product: d/dx(xy) = y + x(dy/dx).
Implicit differentiation of y^n: d/dx(y^n) = n · y^(n-1) · (dy/dx).
Profit = (p - unit cost) × q.
Students often write the Lagrangian as L = f + λg instead of L = f - λg. The sign convention matters; be consistent with whatever your textbook uses. On this exam, L = f - λg is expected (the system works either way, but λ will have the opposite sign).
In implicit differentiation, students forget to attach dy/dx when differentiating y terms. Every time you differentiate a y, the chain rule adds a factor of dy/dx.
For profit problems, students confuse Profit = pq (that is revenue, not profit) with Profit = (p - 5)q. Revenue ignores costs; profit does not.
Students sometimes write Profit = p(q - 5) instead of Profit = (p - 5)q. The $5 is the cost per copy, so it is subtracted from the price per copy, not from the quantity.
⚠️ Lagrange multipliers (Q16) is worth 10 points across three sub-parts. The setup (writing L) is 2 points of easy marks. Do not skip it.
⚠️ Implicit differentiation (Q17) is worth 5 points. Show every step, including the product rule and the chain rule applications.
⚠️ The profit objective function (Q8) is multiple choice, worth 4 points. Read the options carefully. The answer is the one that uses (p - 5), not just p.
Fill in the blank: The Lagrangian for maximising f(x,y) = x²y subject to x + y = 10 is L = x²y - λ(______). (x + y - 10)
True or false: When differentiating 3y² implicitly with respect to x, the result is 6y. (False; it is 6y · dy/dx)
Fill in the blank: If price is p and unit cost is $8, the profit per unit sold is ______. (p - 8)
True or false: In Lagrange multipliers, you solve a system of two equations. (False; three equations: L_x = 0, L_y = 0, and g = 0)
Fill in the blank: To find dy/dx implicitly from x² + y² = 25, differentiate to get 2x + 2y(dy/dx) = 0, so dy/dx = ______. (-x/y)
Q: Use Lagrange multipliers to find the maximum of f(x, y) = 4xy subject to x + 2y = 8.
A: g(x, y) = x + 2y - 8 = 0. L = 4xy - λ(x + 2y - 8). L_x = 4y - λ = 0 → λ = 4y. L_y = 4x - 2λ = 0 → λ = 2x. So 4y = 2x, giving x = 2y. Substitute into constraint: 2y + 2y = 8, so y = 2, x = 4. f(4, 2) = 32.
Q: Find dy/dx for x³ + y³ = 6xy.
A: Differentiate: 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx). Rearrange: 3y²(dy/dx) - 6x(dy/dx) = 6y - 3x². Factor: dy/dx(3y² - 6x) = 6y - 3x². So dy/dx = (6y - 3x²) / (3y² - 6x) = (2y - x²) / (y² - 2x).
Q: A company sells q = 1000 - 50p units at price p dollars. Each unit costs $3 to produce. What is the profit function in terms of p?
A: Profit = (p - 3)q = (p - 3)(1000 - 50p) = 1000p - 50p² - 3000 + 150p = -50p² + 1150p - 3000.
Q: Find dy/dx for e^y + xy = 10.
A: Differentiate: e^y(dy/dx) + y + x(dy/dx) = 0. Factor: dy/dx(e^y + x) = -y. So dy/dx = -y / (e^y + x).
Lagrange multipliers extend the absolute-extrema-on-a-closed-region technique from Doc 2. When the boundary is a curve (not the edges of a rectangle), Lagrange multipliers handle the constraint elegantly.
Implicit differentiation connects to related rates (Doc 1). In related rates, you implicitly differentiate with respect to time t. Here, you do the same with respect to x. The chain rule is the common thread.
The profit setup in Q8 is a gateway to single-variable optimisation: once you have Profit as a function of one variable, you differentiate and set to zero, exactly as in the single-variable techniques from Doc 1.
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