Multivariable Calculus Quiz: Lagrange Multipliers

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Lagrange MultipliersQuestion 1 of 14

Let $P$ be a point on the ellipse $x^2 + 2y^2 = 6$ where the function $f(x,y) = x^2y$ attains its maximum value. Which of the following statements provides the correct geometric interpretation of the relationship between the ellipse and the level curve of $f$ that passes through $P$ ?

The ellipse and the level curve are tangent at

P

The ellipse and the level curve are orthogonal at

P

The gradient of

f

is the zero vector at

P

The point

P

is a vertex of the ellipse.

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Lagrange Multipliers

Practice Lagrange Multipliers in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Lagrange Multipliers, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The ellipse and the level curve are tangent at $P$ . (correct answer)
The ellipse and the level curve are orthogonal at $P$ .
The gradient of $f$ is the zero vector at $P$ .
The point $P$ is a vertex of the ellipse.

Explanation: The method of Lagrange multipliers is based on the principle that at an constrained extremum

P

, the gradient of the function

f

must be parallel to the gradient of the constraint function

g

. That is,

\nabla f(P) = \lambda \nabla g(P)

. The gradient vector

\nabla f

is orthogonal to the level curves of

f

, and

\nabla g

is orthogonal to the level curve of

g

(which is the constraint curve). If the gradient vectors are parallel, their corresponding level curves must be tangent at the point

P

Question 2

A consumer's utility from consuming $x$ units of good A and $y$ units of good B is given by $U(x,y) = xy$ . The price of good A is $2 per unit and the price of good B is $5 per unit. The consumer's budget is $100. Using the method of Lagrange multipliers, the value of the multiplier $\lambda$ represents the marginal utility of money. What is the approximate increase in maximum utility if the consumer's budget increases from $100 to $101?

5 (correct answer)
10
25
250

Explanation: We maximize

U(x,y)=xy

subject to

g(x,y)=2x+5y=100

. The Lagrange system is

\nabla U = \lambda \nabla g

, which is

\langle y, x \rangle = \lambda \langle 2, 5 \rangle

. This yields

y=2\lambda

and

x=5\lambda

. Substituting into the budget constraint:

2(5\lambda) + 5(2\lambda) = 100 \implies 10\lambda + 10\lambda = 100 \implies 20\lambda = 100 \implies \lambda = 5

. The value of

\lambda

approximates the change in the optimal value of

U

for a one-unit increase in the constraint constant. Therefore, if the budget increases by $1, the maximum utility will increase by approximately

\lambda = 5

Question 3

When attempting to find the extrema of $f(x,y) = x+y$ subject to the constraint $g(x,y) = x^3 - y^2 = 0$ , the method of Lagrange multipliers fails to identify the point $(0,0)$ as a potential extremum. What is the reason for this failure?

The function $f$ is not differentiable at $(0,0)$ .
The constraint curve $g(x,y)=0$ is not a closed and bounded set.
The gradient of the constraint function, $\nabla g$ , is the zero vector at $(0,0)$ . (correct answer)
The gradient of the objective function, $\nabla f$ , is the zero vector at $(0,0)$ .

Explanation: When using Lagrange multipliers to find extrema of

f(x,y)

subject to constraint

g(x,y) = 0

, you're looking for points where

\nabla f = \lambda \nabla g

for some scalar

\lambda

. This method requires that

\nabla g \neq \mathbf{0}

at potential extrema—this is called the regularity condition. Let's examine what happens at

(0,0)

. For

g(x,y) = x^3 - y^2

, we have

\nabla g = (3x^2, -2y)

. At

(0,0)

, this gives us

\nabla g(0,0) = (0,0) = \mathbf{0}

. Since the gradient of the constraint function is zero, the Lagrange multiplier equation

\nabla f = \lambda \nabla g

becomes

\nabla f = \lambda \mathbf{0} = \mathbf{0}

. But

\nabla f = (1,1) \neq \mathbf{0}

, so no

\lambda

can satisfy this equation. The method fails because the regularity condition is violated. Answer A is incorrect because

f(x,y) = x + y

is clearly differentiable everywhere, including at

(0,0)

. Answer B misses the point—while the constraint curve

x^3 = y^2

is indeed unbounded, this isn't why Lagrange multipliers fails at this specific point. Answer D is wrong because

\nabla f = (1,1)

is never zero. Remember: Lagrange multipliers can fail when

\nabla g = \mathbf{0}

at a point on the constraint curve. Always check the regularity condition, and when it fails, examine those points separately as potential extrema.

Question 4

The method of Lagrange multipliers is used to find the extrema of a differentiable function $f$ subject to a differentiable constraint $g(x,y,z)=c$ . The method is guaranteed to identify all absolute maximum and minimum values of $f$ under which of the following conditions?

The gradient $\nabla g$ is never the zero vector on the set of points satisfying $g(x,y,z)=c$ .
The set of points satisfying $g(x,y,z)=c$ is closed and bounded.
The set of points satisfying $g(x,y,z)=c$ is closed and bounded, and $\nabla g \neq \mathbf{0}$ on this set. (correct answer)
The function $f$ is linear and the constraint function $g$ is quadratic.

Explanation: When applying Lagrange multipliers to find extrema of a function

f

subject to constraint

g(x,y,z)=c

, you're looking for points where

\nabla f = \lambda \nabla g

. However, this method alone doesn't guarantee you'll find all absolute extrema—you need additional conditions to ensure completeness. The correct answer is C because it combines two essential requirements. First, the constraint set must be closed and bounded (compact), which guarantees by the Extreme Value Theorem that continuous function

f

will attain its absolute maximum and minimum on this set. Second,

\nabla g \neq \mathbf{0}

everywhere on the constraint set ensures the constraint qualifies as a "regular" curve/surface where the Lagrange multiplier method is valid—this prevents degenerate cases where the method might miss critical points. Option A gives you the regularity condition but lacks compactness, so extrema might not exist (imagine optimizing on an unbounded curve). Option B provides compactness but without

\nabla g \neq \mathbf{0}

, you could have singular points where Lagrange multipliers fail—like trying to optimize at a cusp or corner where the constraint isn't smooth. Option D incorrectly suggests that specific function types (linear

f

, quadratic

g

) alone guarantee the method works, but the geometric properties of the constraint set matter more than the algebraic forms. Study tip: Remember that Lagrange multipliers need both analytical conditions (

\nabla g \neq \mathbf{0}

) and topological conditions (closed, bounded constraint set) to guarantee finding all absolute extrema. Always check both requirements.

Question 5

A manufacturer's production is modeled by the Cobb-Douglas function $P(K, L) = 100 K^{1/4} L^{3/4}$ , where $K$ is units of capital and $L$ is units of labor. The budget for production is $120,000, with capital costing $300 per unit and labor costing $100 per unit. What is the maximum production level $P$ ?

30,000
40,000
$25\sqrt{3}$
$30000\sqrt{3}$ (correct answer)

Explanation: This is a constrained optimization problem where you need to maximize a production function subject to a budget constraint. When you see a Cobb-Douglas function with a linear budget constraint, think Lagrange multipliers or the method of substitution. Set up the constraint equation:

300K + 100L = 120,000

, which simplifies to

3K + L = 1200

. From this,

L = 1200 - 3K

. Substitute into the production function:

P(K) = 100K^{1/4}(1200-3K)^{3/4}

. To find the maximum, take the derivative and set it equal to zero:

P'(K) = 100 \left[\frac{1}{4}K^{-3/4}(1200-3K)^{3/4} + K^{1/4} \cdot \frac{3}{4}(1200-3K)^{-1/4} \cdot (-3)\right] = 0

Factoring out common terms and simplifying:

\frac{1200-3K}{4K} = \frac{9}{4}

Solving:

1200-3K = 9K

, so

K = 100

and

L = 900

. The maximum production is:

P(100,900) = 100 \cdot 100^{1/4} \cdot 900^{3/4} = 100 \cdot \sqrt[4]{100} \cdot \sqrt[4]{900^3}

Since

100^{1/4} = \sqrt{10}

and

900^{3/4} = (30^2)^{3/4} = 30^{3/2} = 30\sqrt{30} = 30\sqrt{30}

P = 100 \cdot \sqrt{10} \cdot 30\sqrt{30} = 30000\sqrt{3}

Option A (30,000) ignores the irrational component. Option B (40,000) likely comes from computational errors. Option C (

25\sqrt{3}

) has the right irrational form but wrong magnitude, probably from early algebraic mistakes. Study tip: In Cobb-Douglas optimization problems, always double-check your arithmetic with radicals—the final answer often involves square roots that students accidentally drop or miscalculate.

Question 6

A particle moves along the curve $x^2 + 2y^2 = 6$ . At which point on this curve is the function $h(x,y) = x^2y$ maximized?

$(2, 1)$ (correct answer)
$(\sqrt{2}, \sqrt{2})$
$(2, -1)$
$(\sqrt{6}, 0)$

Explanation: Using Lagrange multipliers with constraint

g(x,y) = x^2 + 2y^2 - 6 = 0

. We need

\nabla h = \lambda \nabla g

, giving

(2xy, x^2) = \lambda(2x, 4y)

. This yields

2xy = 2\lambda x

and

x^2 = 4\lambda y

. From the first equation (assuming

x \neq 0

y = \lambda

. Substituting into the second:

x^2 = 4y^2

, so

x = \pm 2y

. Using the constraint

x^2 + 2y^2 = 6

: when

x = 2y

, we get

4y^2 + 2y^2 = 6

, so

6y^2 = 6

, giving

y = \pm 1

. When

x = -2y

, we get the same

y

values. This gives us critical points:

(2,1)

(2,-1)

(-2,1)

(-2,-1)

. We also need to check the case

x = 0

: from the constraint,

2y^2 = 6

, so

y = \pm\sqrt{3}

, giving points

(0, \sqrt{3})

and

(0, -\sqrt{3})

. Evaluating

h(x,y) = x^2y

: At

(2,1)

h = 4(1) = 4

. At

(2,-1)

h = 4(-1) = -4

. At

(-2,1)

h = 4(1) = 4

. At

(-2,-1)

h = 4(-1) = -4

. At

(0, \pm\sqrt{3})

h = 0

. The maximum value is 4, achieved at

(2,1)

and

(-2,1)

. Since only

(2,1)

appears in the options, this is the answer. Choice B doesn't satisfy the constraint. Choice C gives a minimum. Choice D gives

h = 0

Question 7

What is the minimum value of the function $S(x,y) = 2x+8y$ for points $(x,y)$ in the first quadrant on the hyperbola $xy=9$ ?

12
24 (correct answer)
26
30

Explanation: We want to minimize

f(x,y) = 2x+8y

subject to

g(x,y) = xy-9=0

with

x,y>0

. The Lagrange system is

\nabla f = \lambda \nabla g

, which gives

\langle 2, 8 \rangle = \lambda \langle y, x \rangle

. This yields the equations

2=\lambda y

and

8=\lambda x

. Since

x,y>0

\lambda

must be non-zero. From these equations,

y = 2/\lambda

and

x=8/\lambda

, which implies

x=4y

. Substituting this into the constraint:

(4y)y = 9 \implies 4y^2 = 9 \implies y=3/2

(since we are in the first quadrant). Then

x=4(3/2)=6

. The minimum value occurs at

(6, 3/2)

and is

S(6, 3/2) = 2(6) + 8(3/2) = 12 + 12 = 24

. Distractor A comes from an error in applying the AM-GM inequality. Distractors C and D come from testing simple integer points on the hyperbola like

(9,1)

and

(3,3)

Question 8

Find the maximum value of the function $f(x, y) = 4x + 3y$ subject to the constraint given by the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$ .

12
24 (correct answer)
30
48

Explanation: Let

g(x,y) = \frac{x^2}{18} + \frac{y^2}{32} - 1

. We set up the Lagrange multiplier system

\nabla f = \lambda \nabla g

. This gives

\langle 4, 3 \rangle = \lambda \langle \frac{x}{9}, \frac{y}{16} \rangle

. We have the equations

4 = \lambda \frac{x}{9}

and

3 = \lambda \frac{y}{16}

. Solving for

x

and

y

gives

x = \frac{36}{\lambda}

and

y = \frac{48}{\lambda}

. Substituting these into the constraint equation:

\frac{(36/\lambda)^2}{18} + \frac{(48/\lambda)^2}{32} = 1

. This simplifies to

\frac{1296}{18\lambda^2} + \frac{2304}{32\lambda^2} = 1

, which becomes

\frac{72}{\lambda^2} + \frac{72}{\lambda^2} = 1

, so

\frac{144}{\lambda^2} = 1

. Thus,

\lambda^2 = 144

, and

\lambda = \pm 12

. For

\lambda = 12

, we get

(x,y) = (3,4)

, and

f(3,4) = 4(3) + 3(4) = 24

. For

\lambda = -12

, we get

(x,y) = (-3,-4)

, and

f(-3,-4) = -24

. The maximum value is 24.

Question 9

Let $P=(x,y,z)$ be the point on the sphere $x^2 + y^2 + z^2 = 84$ that is closest to the point $(1, 2, 3)$ . What is the value of the sum of the coordinates, $x+y+z$ ?

$6\sqrt{6}$ (correct answer)
$-6\sqrt{6}$
14
$\sqrt{84}$

Explanation: We want to minimize the distance, which is equivalent to minimizing the squared distance function

f(x,y,z) = (x-1)^2 + (y-2)^2 + (z-3)^2

subject to the constraint

g(x,y,z) = x^2 + y^2 + z^2 = 84

. The Lagrange system is

\nabla f = \lambda \nabla g

, which gives

\langle 2(x-1), 2(y-2), 2(z-3) \rangle = \lambda \langle 2x, 2y, 2z \rangle

. This leads to

x-1=\lambda x

y-2=\lambda y

, and

z-3=\lambda z

. Solving for

x, y, z

gives

x=\frac{1}{1-\lambda}

y=\frac{2}{1-\lambda}

z=\frac{3}{1-\lambda}

. This implies

y=2x

and

z=3x

. Substituting into the constraint:

x^2 + (2x)^2 + (3x)^2 = 84 \implies 14x^2 = 84 \implies x^2 = 6 \implies x = \pm\sqrt{6}

. The two candidate points are

(\sqrt{6}, 2\sqrt{6}, 3\sqrt{6})

and

(-\sqrt{6}, -2\sqrt{6}, -3\sqrt{6})

. The point

(\sqrt{6}, 2\sqrt{6}, 3\sqrt{6})

is closer to

(1,2,3)

. The sum of its coordinates is

\sqrt{6} + 2\sqrt{6} + 3\sqrt{6} = 6\sqrt{6}

Question 10

Find the maximum value of the function $f(x,y,z) = x - 2y + 2z$ for points on the curve formed by the intersection of the cylinder $x^2 + y^2 = 1$ and the plane $x+y+z=1$ .

3
$2 - \sqrt{17}$
$2 + \sqrt{17}$ (correct answer)
1

Explanation: This is a constrained optimization problem where you need to find the maximum of a linear function subject to two constraints simultaneously. When you see this setup, think Lagrange multipliers - you're optimizing over the intersection of two surfaces. The curve is defined by two constraints:

g_1(x,y,z) = x^2 + y^2 - 1 = 0

(cylinder) and

g_2(x,y,z) = x + y + z - 1 = 0

(plane). Using Lagrange multipliers, you need

\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2

. This gives you:

\nabla f = (1, -2, 2)

\nabla g_1 = (2x, 2y, 0)

, and

\nabla g_2 = (1, 1, 1)

. Setting up the system:

$1 = 2\lambda_1 x + \lambda_2$
$-2 = 2\lambda_1 y + \lambda_2$
$2 = \lambda_2$

From the third equation,

\lambda_2 = 2

. Substituting back:

2\lambda_1 x = -1

and

2\lambda_1 y = -4

, so

x = -\frac{1}{2\lambda_1}

and

y = -\frac{2}{\lambda_1}

. Since these points must satisfy

x^2 + y^2 = 1

, you get

\lambda_1 = \pm\frac{\sqrt{17}}{2}

. For maximum value, use

\lambda_1 = -\frac{\sqrt{17}}{2}

, giving

(x,y) = \left(\frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right)

. From the plane constraint,

z = 1 - x - y = 1 - \frac{5}{\sqrt{17}}

. Thus

f = \frac{1}{\sqrt{17}} - \frac{8}{\sqrt{17}} + 2 - \frac{10}{\sqrt{17}} = 2 + \sqrt{17}

. Choice (A) 3 likely comes from ignoring one constraint. Choice (B)

2 - \sqrt{17}

is the minimum value. Choice (D) 1 might result from computational errors. Remember: constrained optimization with multiple constraints requires Lagrange multipliers, and always check whether you found a maximum or minimum by examining the sign of your multipliers.

Question 11

An open-topped rectangular box with a square base is to be constructed with a volume of exactly 32 cubic meters. What is the minimum possible surface area of the material needed for its construction?

32 m $^2$
48 m $^2$ (correct answer)
$48\sqrt[3]{2}$ m $^2$
54 m $^2$

Explanation: Let the square base have side length

x

and the box have height

y

. The volume is

V = x^2y = 32

. This is the constraint. The surface area of the open-topped box is

S = x^2 + 4xy

. We want to minimize

S

. Let

f(x,y) = x^2+4xy

and

g(x,y) = x^2y - 32

. The system

\nabla f = \lambda \nabla g

gives

\langle 2x+4y, 4x \rangle = \lambda \langle 2xy, x^2 \rangle

. From the second component,

4x = \lambda x^2

, which implies

\lambda = 4/x

(since

x>0

). Substituting into the first component equation:

2x+4y = (4/x)(2xy) = 8y

. This simplifies to

2x = 4y

, or

x=2y

. Substitute this into the volume constraint:

(2y)^2 y = 32 \implies 4y^3 = 32 \implies y^3 = 8 \implies y=2

. Then

x = 2(2) = 4

. The minimum surface area is

S = (4)^2 + 4(4)(2) = 16 + 32 = 48

^2

Question 12

What is the absolute minimum value of the function $f(x,y) = x^2 - y^2$ subject to the constraint $x^2 + y^2 = 4$ ?

4
0
-2
-4 (correct answer)

Explanation: Let

g(x,y) = x^2+y^2-4=0

. The Lagrange system is

\nabla f = \lambda \nabla g

, which gives

\langle 2x, -2y \rangle = \lambda \langle 2x, 2y \rangle

. This results in two equations: (1)

2x = 2\lambda x

and (2)

-2y = 2\lambda y

. From (1),

2x(1-\lambda)=0

, so either

x=0

\lambda=1

. From (2),

2y(1+\lambda)=0

, so either

y=0

\lambda=-1

. We examine the cases: Case 1:

x=0

. The constraint gives

y^2=4

, so

y=\pm 2

. The points are

(0, 2)

and

(0, -2)

. For both points,

f(x,y) = 0^2 - (\pm 2)^2 = -4

. Case 2:

y=0

. The constraint gives

x^2=4

, so

x=\pm 2

. The points are

(2, 0)

and

(-2, 0)

. For both points,

f(x,y) = (\pm 2)^2 - 0^2 = 4

. We cannot have

\lambda=1

and

\lambda=-1

simultaneously. Comparing the function values, the absolute minimum is -4.

Question 13

Find the maximum value of $f(x,y,z) = x + 2y + 3z$ subject to the constraints $x^2 + y^2 + z^2 = 14$ and $x + y + z = 0$ .

$2\sqrt{6}$
$2\sqrt{7}$ (correct answer)
$3\sqrt{2}$
$\sqrt{42}$

Explanation: This requires two Lagrange multipliers since we have two constraints. Let

g_1(x,y,z) = x^2 + y^2 + z^2 - 14

and

g_2(x,y,z) = x + y + z

. The conditions are:

\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2

, giving us

1 = 2\lambda_1 x + \lambda_2

2 = 2\lambda_1 y + \lambda_2

3 = 2\lambda_1 z + \lambda_2

. From these:

x = \frac{1-\lambda_2}{2\lambda_1}

y = \frac{2-\lambda_2}{2\lambda_1}

z = \frac{3-\lambda_2}{2\lambda_1}

. Using the constraint

x + y + z = 0

\frac{(1-\lambda_2) + (2-\lambda_2) + (3-\lambda_2)}{2\lambda_1} = 0

, so

6 - 3\lambda_2 = 0

, giving

\lambda_2 = 2

. Thus

x = \frac{-1}{2\lambda_1}

y = 0

z = \frac{1}{2\lambda_1}

. Using

x^2 + y^2 + z^2 = 14

\frac{1}{4\lambda_1^2} + 0 + \frac{1}{4\lambda_1^2} = 14

, so

\frac{1}{2\lambda_1^2} = 14

, giving

\lambda_1^2 = \frac{1}{28}

, so

\lambda_1 = \pm\frac{1}{2\sqrt{7}}

. For maximum, we want

\lambda_1 = -\frac{1}{2\sqrt{7}}

, giving

x = \sqrt{7}

y = 0

z = -\sqrt{7}

. The maximum value is

\sqrt{7} + 0 + 3(-\sqrt{7}) = \sqrt{7} - 3\sqrt{7} = -2\sqrt{7}

. Wait, this is negative. For maximum, we want

\lambda_1 = \frac{1}{2\sqrt{7}}

, giving

x = -\sqrt{7}

y = 0

z = \sqrt{7}

. The maximum value is

-\sqrt{7} + 0 + 3\sqrt{7} = 2\sqrt{7}

. Choice A comes from computational errors. Choice C results from incorrect constraint handling. Choice D represents incorrect combination of the constraint values.

Question 14

Consider the optimization problem: maximize $f(x,y) = xy$ subject to $x^2 + y^2 = 8$ . Which statement about the critical points is correct?

There are exactly two critical points, both giving local maxima with function value 4
There are exactly four critical points, with two giving local maxima with function value 4 (correct answer)
There are exactly four critical points, with two giving local maxima with function value 2
There are exactly two critical points, both giving local maxima with function value 2

Explanation: Using Lagrange multipliers:

\nabla f = \lambda \nabla g

where

g(x,y) = x^2 + y^2 - 8

. This gives

y = 2\lambda x

and

x = 2\lambda y

. Substituting the first into the second:

x = 2\lambda(2\lambda x) = 4\lambda^2 x

. For

x \neq 0

, we get

1 = 4\lambda^2

, so

\lambda = \pm\frac{1}{2}

. When

\lambda = \frac{1}{2}

y = x

, and from the constraint

2x^2 = 8

, so

x = \pm 2

, giving points

(2,2)

and

(-2,-2)

with

f = 4

. When

\lambda = -\frac{1}{2}

y = -x

, giving points

(2,-2)

and

(-2,2)

with

f = -4

. The points with

f = 4

are local maxima. Choice A misses half the critical points. Choice C uses incorrect constraint radius. Choice D both misses critical points and has wrong function value.