Multivariable Calculus Quiz: Constrained Optimization

15 questions · exam conditions

0:00

Constrained OptimizationQuestion 1 of 15

An environmental sensor measures pollution concentration $P(x,y) = 100 - (x-5)^2 - (y-5)^2$ at location $(x,y)$ . The sensor can only operate in a region $R$ defined by $x \ge 0$ , $y \ge 0$ , and $x+2y \ge 16$ . What is the maximum pollution concentration the sensor can record in the allowed region?

66

99

99.8

100

← Back to quizzes

Multivariable Calculus Quiz

Multivariable Calculus Quiz: Constrained Optimization

Practice Constrained Optimization 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 Constrained Optimization, 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

$66$
$99$
$99.8$ (correct answer)
$100$

Explanation: The function

P(x,y)

is a downward-opening paraboloid with its unconstrained maximum of

100

(5,5)

. We check if this point is in the allowed region

R

. While

5 \ge 0

and

5 \ge 0

, the third condition

5 + 2(5) = 15 \ge 16

is not satisfied. Thus, the maximum must occur on the boundary of

R

. Since the level curves of

P

are circles centered at

(5,5)

, the maximum value on

R

will occur at the point in

R

that is closest to

(5,5)

. This point must lie on the line

x+2y=16

. We minimize the squared distance

d^2 = (x-5)^2 + (y-5)^2

subject to

x+2y=16

. Using Lagrange multipliers,

\langle 2(x-5), 2(y-5) \rangle = \lambda \langle 1, 2 \rangle

. This gives

2(x-5)=\lambda

and

2(y-5)=2\lambda

, which implies

y-5 = 2(x-5)

, or

y=2x-5

. Substituting into the constraint:

x+2(2x-5)=16 \implies 5x-10=16 \implies x=26/5=5.2

. Then

y=2(5.2)-5=5.4

. At this point

(5.2, 5.4)

, the pollution is

P(5.2, 5.4) = 100 - (5.2-5)^2 - (5.4-5)^2 = 100 - 0.04 - 0.16 = 99.8

. Checking the corners of the boundary,

P(16,0) = -46

and

P(0,8)=66

, confirms that

99.8

is the maximum.

Question 2

Find the point on the plane $x+y+z=6$ that is closest to the point $P(1,1,1)$ . Let this closest point be $Q(x_0, y_0, z_0)$ . What is the value of $x_0+y_0$ ?

$2$
$3$
$4$ (correct answer)
$5$

Explanation: We want to find the point

Q(x_0, y_0, z_0)

on the plane that minimizes the distance to

P(1,1,1)

. This is equivalent to minimizing the squared distance

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

subject to the constraint

g(x,y,z) = x+y+z-6=0

. The vector from

Q

P

\vec{QP} = \langle 1-x, 1-y, 1-z \rangle

, must be parallel to the plane's normal vector,

\vec{n}=\langle 1,1,1 \rangle

. Thus, there is a scalar

k

such that

\vec{QP} = k\vec{n}

. This gives

1-x=k

1-y=k

, and

1-z=k

. So

x=1-k

y=1-k

, and

z=1-k

. Since

Q

is on the plane, its coordinates must sum to 6:

(1-k) + (1-k) + (1-k) = 6 \implies 3-3k=6 \implies -3k=3 \implies k=-1

. The coordinates of the point

Q

are

x_0 = 1-(-1)=2

y_0=1-(-1)=2

, and

z_0=1-(-1)=2

. The question asks for the value of

x_0+y_0

, which is

2+2=4

Question 3

The function $f(x,y) = xy$ is to be maximized subject to the constraint $(x-1)^2 + y^2 = 1$ . The Lagrange multiplier method yields two critical points. One point gives the maximum value and the other gives the minimum value. What is the minimum value?

$-\frac{\sqrt{3}}{2}$
$0$
$-1$
$-\frac{3\sqrt{3}}{4}$ (correct answer)

Explanation: When you encounter a constrained optimization problem like this, you're dealing with the method of Lagrange multipliers, which finds extrema of a function subject to a constraint by solving

\nabla f = \lambda \nabla g

where

g(x,y) = 0

is your constraint. Here, you want to optimize

f(x,y) = xy

subject to

(x-1)^2 + y^2 = 1

(a circle centered at

(1,0)

with radius 1). Setting up the Lagrangian:

\nabla f = (y,x)

and

\nabla g = (2(x-1), 2y)

, so you need

y = 2\lambda(x-1)

and

x = 2\lambda y

. From the second equation,

\lambda = \frac{x}{2y}

(assuming

y \neq 0

). Substituting into the first:

y = 2 \cdot \frac{x}{2y} \cdot (x-1) = \frac{x(x-1)}{y}

, which gives

y^2 = x(x-1)

. Using the constraint

(x-1)^2 + y^2 = 1

and substituting:

(x-1)^2 + x(x-1) = 1

. Let

u = x-1

, so

u^2 + (u+1)u = 1

, which simplifies to

2u^2 + u - 1 = 0

. This factors as

(2u-1)(u+1) = 0

, giving

u = \frac{1}{2}

u = -1

. This yields

x = \frac{3}{2}, y = \pm\frac{\sqrt{3}}{2}

and

x = 0, y = 0

. The function values are

f\left(\frac{3}{2}, \pm\frac{\sqrt{3}}{2}\right) = \pm\frac{3\sqrt{3}}{4}

and

f(0,0) = 0

. The minimum is

-\frac{3\sqrt{3}}{4}

(choice D). Choice A represents a calculation error, choice B is the value at one critical point but not the minimum, and choice C likely comes from misunderstanding the constraint geometry. Strategy tip: Always evaluate your function at all critical points found via Lagrange multipliers to identify which gives the maximum versus minimum.

Question 4

A factory's daily production is modeled by the Cobb-Douglas function $Q(K, L) = 60 K^{1/3} L^{2/3}$ , where $K$ is the capital investment in units of $1,000 and $L$ is the labor input in worker-hours. The budget constraint is given by $10K + 20L = 6000$ . For the values of $K$ and $L$ that maximize production, the Lagrange multiplier $\lambda$ represents the marginal productivity of money (the rate of change of maximum production with respect to the budget). What is the approximate increase in maximum production if the budget is increased by one dollar?

$0.5$
$2$ (correct answer)
$200$
$12000$

Explanation: We want to maximize

Q(K, L)

subject to the constraint

g(K, L) = 10K + 20L = 6000

. We set

\nabla Q = \lambda \nabla g

. The gradients are

\nabla Q = \langle 20(L/K)^{2/3}, 40(K/L)^{1/3} \rangle

and

\nabla g = \langle 10, 20 \rangle

. This gives the system:

20(L/K)^{2/3} = 10\lambda

and

40(K/L)^{1/3} = 20\lambda

. Dividing the second equation by the first gives

2(K/L) = 2

, which simplifies to

K=L

. Substituting this into the budget constraint:

10K + 20K = 6000

, which gives

30K=6000

, so

K=200

and

L=200

. To find

\lambda

, we substitute these values back into one of the gradient equations:

40(200/200)^{1/3} = 20\lambda

, which simplifies to

40(1) = 20\lambda

, so

\lambda=2

. The value of

\lambda

is the marginal productivity of money, so an increase of $1 in the budget results in an approximate increase of 2 units in production.

Question 5

To find the extreme values of $f(x,y,z) = x^2+y^2+z^2$ subject to the constraint $g(x,y,z) = x^2+y^2-z^2=0$ , a student sets up the Lagrange multiplier system $\nabla f = \lambda \nabla g$ . This leads to the equations $2x=\lambda(2x)$ , $2y=\lambda(2y)$ , and $2z=\lambda(-2z)$ . Which of the following describes the set of points satisfying this system and the constraint?

The system has no solution.
Only the point $(0,0,0)$ . (correct answer)
The points on the circle $x^2+y^2=1$ in the plane $z=1$ .
The point $(0,0,0)$ and the points on the circle $x^2+y^2=1$ in the plane $z=-1$ .

Explanation: The Lagrange multiplier system is: (1)

2x(1-\lambda)=0

, (2)

2y(1-\lambda)=0

, (3)

2z(1+\lambda)=0

. From (1), either

x=0

\lambda=1

. From (2), either

y=0

\lambda=1

. From (3), either

z=0

\lambda=-1

. Case 1:

\lambda=1

. Then equation (3) becomes

2z(2)=0

, so

z=0

. The constraint is

x^2+y^2-z^2=0

, which becomes

x^2+y^2=0

. This implies

x=0

and

y=0

. So we get the point

(0,0,0)

. Case 2:

\lambda=-1

. Then equations (1) and (2) become

2x(2)=0

and

2y(2)=0

, which implies

x=0

and

y=0

. The constraint

x^2+y^2-z^2=0

gives

0-z^2=0

, so

z=0

. Again, we get the point

(0,0,0)

. Case 3:

\lambda \neq 1

and

\lambda \neq -1

. Then

x=0

y=0

, and

z=0

. So all cases lead to the single point

(0,0,0)

. This point satisfies the constraint

0^2+0^2-0^2=0

. Therefore,

(0,0,0)

is the only point that satisfies the system. The method of Lagrange multipliers requires

\nabla g \neq \vec{0}

. At

(0,0,0)

\nabla g = \langle 2x, 2y, -2z \rangle = \langle 0,0,0 \rangle

. The method technically fails at this point, but it's the only solution candidate derived from the algebraic system.

Question 6

Consider the problem of finding the extreme values of a differentiable function $f(x,y)$ subject to a constraint $g(x,y)=c$ . The method of Lagrange multipliers is applied, yielding several candidate points $(x_i, y_i)$ . Under which of the following conditions is the method guaranteed to have found the absolute maximum and minimum values of $f$ ?

The function $f(x,y)$ has a second derivative that is always negative.
The Lagrange multiplier $\lambda$ has the same sign at all candidate points.
The gradient $\nabla g$ is never the zero vector at any candidate point.
The constraint curve $g(x,y)=c$ is a closed and bounded set. (correct answer)

Explanation: When you encounter Lagrange multiplier problems asking about guaranteed absolute extrema, focus on the compactness of the constraint set. The method of Lagrange multipliers finds critical points, but additional conditions are needed to ensure these include absolute maximum and minimum values. The correct answer is D because of the Extreme Value Theorem. When the constraint curve

g(x,y) = c

forms a closed and bounded set, the continuous function

f

restricted to this constraint must attain its absolute maximum and minimum values somewhere on the curve. Since Lagrange multipliers finds all critical points where extrema can occur on smooth constraint curves, the method is guaranteed to locate these absolute extrema among its candidate points. Option A is incorrect because the second derivative condition for

f

(concavity) doesn't guarantee that absolute extrema exist on the constraint set—it only tells you about the nature of critical points if they exist. Option B misunderstands the role of

\lambda

; the sign of the Lagrange multiplier indicates whether you're at a local maximum or minimum relative to the constraint, but having the same sign at all points doesn't guarantee absolute extrema exist. Option C identifies a necessary condition for the method to work properly (avoiding degenerate cases), but this alone doesn't ensure absolute extrema exist. Remember: Lagrange multipliers finds candidates for extrema, but you need the constraint set to be compact (closed and bounded) to guarantee that absolute maximum and minimum values actually exist for the continuous function to find.

Question 7

A particle is constrained to move on the surface $x^2 + y^2 + z^2 = 25$ . The temperature at any point is given by $T(x,y,z) = x^2 + y^2 - z^2$ . What is the minimum temperature experienced by the particle?

$-25$ (correct answer)
$-50$
$0$
$25$

Explanation: We want to minimize

T(x,y,z) = x^2 + y^2 - z^2

subject to

x^2 + y^2 + z^2 = 25

. Using the constraint, we can write

x^2 + y^2 = 25 - z^2

, so

T = (25 - z^2) - z^2 = 25 - 2z^2

. Since we're on the sphere,

-5 \leq z \leq 5

. To minimize

T

, we need to maximize

z^2

, which occurs when

z = \pm 5

. When

z = \pm 5

, we have

x^2 + y^2 = 0

, so

x = y = 0

. The minimum temperature is

T(0,0,\pm 5) = 0 + 0 - 25 = -25

Question 8

Consider the optimization problem: maximize $f(x,y,z) = xyz$ subject to $x + 2y + 3z = 12$ where $x,y,z > 0$ . At the optimal point, what is the ratio $x:y:z$ ?

$1:1:1$
$6:3:2$ (correct answer)
$3:3:2$
$2:2:1$

Explanation: Using Lagrange multipliers:

\nabla f = \lambda \nabla g

where

g(x,y,z) = x + 2y + 3z - 12

. This gives us

yz = \lambda

xz = 2\lambda

, and

xy = 3\lambda

. From these equations:

\frac{xz}{yz} = \frac{2\lambda}{\lambda} = 2

, so

x = 2y

. Similarly,

\frac{xy}{yz} = \frac{3\lambda}{\lambda} = 3

, so

x = 3z

. Therefore,

x = 2y = 3z

. Let

z = t

, then

y = \frac{3t}{2}

and

x = 3t

. Substituting into the constraint:

3t + 2(\frac{3t}{2}) + 3t = 12

, so

3t + 3t + 3t = 9t = 12

, giving

t = \frac{4}{3}

. Thus

z = \frac{4}{3}

y = 2

, and

x = 4

. The ratio is

x:y:z = 4:2:\frac{4}{3} = 6:3:2

Question 9

Find the maximum value of $f(x,y) = xy$ subject to the constraint $x^2 + 4y^2 = 8$ where $x > 0$ and $y > 0$ .

$1$
$2$ (correct answer)
$4$
$\sqrt{2}$

Explanation: Using Lagrange multipliers, we need

\nabla f = \lambda \nabla g

where

g(x,y) = x^2 + 4y^2 - 8

. This gives us

y = \lambda \cdot 2x

and

x = \lambda \cdot 8y

. From the first equation,

\lambda = y/(2x)

. Substituting into the second:

x = \frac{y}{2x} \cdot 8y = \frac{4y^2}{x}

, so

x^2 = 4y^2

. Since

x, y > 0

, we have

x = 2y

. Substituting into the constraint:

(2y)^2 + 4y^2 = 8

, so

8y^2 = 8

, giving

y = 1

and

x = 2

. Therefore, the maximum value is

f(2,1) = 2 \cdot 1 = 2

Question 10

What is the absolute maximum value of the function $f(x, y) = 2x^2 + y^2 - y$ on the disk $D = \{(x, y) | x^2 + y^2 \leq 1\}$ ?

$9/4$ (correct answer)
$2$
$0$
$-1/4$

Explanation: First, we find critical points in the interior of the disk.

\nabla f = \langle 4x, 2y-1 \rangle = \langle 0, 0 \rangle

yields the point

(0, 1/2)

. Since

0^2 + (1/2)^2 = 1/4 < 1

, this point is in the interior. The value is

f(0, 1/2) = (1/2)^2 - 1/2 = -1/4

. Next, we find extrema on the boundary

x^2+y^2=1

. We can substitute

x^2 = 1-y^2

into

f

to get a function of

y

h(y) = 2(1-y^2) + y^2 - y = -y^2 - y + 2

, for

y \in [-1, 1]

. To find the maximum of this quadratic on the interval, we check the critical point and the endpoints.

h'(y) = -2y-1 = 0

gives

y = -1/2

. The value is

h(-1/2) = -(-1/2)^2 - (-1/2) + 2 = -1/4+1/2+2 = 9/4

. At the endpoints,

h(-1) = -1 - (-1) + 2 = 2

and

h(1) = -1 - 1 + 2 = 0

. Comparing all candidate values (

-1/4, 9/4, 2, 0

), the absolute maximum is

9/4

Question 11

Find the maximum value of $f(x, y, z) = x$ on the intersection of the sphere $x^2 + y^2 + z^2 = 36$ and the plane $x - 2y + 2z = 18$ .

$2$ (correct answer)
$4$
$6$
No maximum exists because the intersection is empty.

Explanation: First, determine the nature of the intersection. The constraint is the intersection of a sphere centered at the origin with radius

R=6

and a plane. The distance from the sphere's center

(0,0,0)

to the plane

x - 2y + 2z - 18 = 0

d = |1(0) - 2(0) + 2(0) - 18| / \sqrt{1^2 + (-2)^2 + 2^2} = |-18| / \sqrt{9} = 18/3 = 6

. Since the distance from the center to the plane is equal to the radius of the sphere, the plane is tangent to the sphere at a single point. Therefore, the constraint set consists of only one point. The function

f(x, y, z) = x

is evaluated at this single point. The point of tangency is found by moving from the origin along the plane's normal vector

\vec{n}=\langle 1, -2, 2 \rangle

by a distance of the radius, 6. The unit normal is

\hat{n} = \langle 1/3, -2/3, 2/3 \rangle

. The point is

6 \hat{n} = (2, -4, 4)

. Alternatively, solve for the point of tangency by substituting

x=t, y=-2t, z=2t

into the plane equation:

t-2(-2t)+2(2t)=18 \implies 9t=18 \implies t=2

. So the point is

(2,-4,4)

. For this single point, the value of

f(x,y,z)=x

is 2. This is trivially the maximum and minimum value. Distractor D is incorrect because the intersection is not empty.

Question 12

A rectangular box with no top is to be constructed from 12 square meters of cardboard. What is the maximum possible volume of such a box?

$2 \text{ m}^3$
$4 \text{ m}^3$ (correct answer)
$6 \text{ m}^3$
$8 \text{ m}^3$

Explanation: Let the dimensions of the base be

x

and

y

, and the height be

z

. The volume is

V = xyz

. The surface area of the base is

xy

and the area of the four sides is

2xz + 2yz

. Since there is no top, the total surface area is

A = xy + 2xz + 2yz = 12

. We want to maximize

V(x,y,z) = xyz

subject to this constraint. Using Lagrange multipliers,

\nabla V = \lambda \nabla A

. This gives

\langle yz, xz, xy \rangle = \lambda \langle y+2z, x+2z, 2x+2y \rangle

. The system is: (1)

yz = \lambda(y+2z)

, (2)

xz = \lambda(x+2z)

, (3)

xy = \lambda(2x+2y)

. From (1) and (2), we can deduce

\frac{yz}{y+2z} = \frac{xz}{x+2z}

(assuming non-zero denominators). Since

z \neq 0

for a box,

y(x+2z) = x(y+2z)

, which simplifies to

xy+2yz = xy+2xz

, so

2yz=2xz

. Since

z \neq 0

, we have

x=y

. Substitute

x=y

into (3):

x^2 = \lambda(4x)

. Since

x \neq 0

x=4\lambda

. Substitute

x=y

into (1):

xz = \lambda(x+2z)

. Substitute

x=4\lambda

(4\lambda)z = \lambda(4\lambda+2z)

. Since

\lambda \neq 0

4z = 4\lambda+2z

, which means

2z = 4\lambda

, so

z=2\lambda

. Thus, we have

x=y=4\lambda

and

z=2\lambda

, which means

x=y=2z

. Substitute this into the area constraint:

x(x) + 2x(x/2) + 2x(x/2) = 12 \implies x^2 + x^2 + x^2 = 12 \implies 3x^2 = 12 \implies x^2=4 \implies x=2

. Then

y=2

and

z=1

. The maximum volume is

V = (2)(2)(1) = 4 \text{ m}^3

Question 13

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

$1$
$4 - 2\sqrt{2}$
$4 + 2\sqrt{2}$ (correct answer)
$1 + \sqrt{2}$

Explanation: We use Lagrange multipliers with two constraints,

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

and

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

. The system is

\nabla f = \lambda \nabla g + \mu \nabla h

, which gives:

2x = \lambda(2x) + \mu(1)

2y = \lambda(2y) + \mu(1)

, and

2z = \mu(1)

. From the first two equations,

2x(1-\lambda) = \mu

and

2y(1-\lambda)=\mu

. If

\lambda \neq 1

, then

x=y

. Substituting into

x^2+y^2=1

gives

2x^2=1

, so

x=y=\pm 1/\sqrt{2}

. From

x+y+z=1

, we find two points:

(1/\sqrt{2}, 1/\sqrt{2}, 1-\sqrt{2})

and

(-1/\sqrt{2}, -1/\sqrt{2}, 1+\sqrt{2})

. If

\lambda = 1

, then

\mu=0

, which implies

z=0

. Then

x+y=1

and

x^2+y^2=1

, giving points

(1,0,0)

and

(0,1,0)

. We evaluate

f

at these four candidate points. Note

f = x^2+y^2+z^2 = 1+z^2

. For the first two points,

f = 1+(1-\sqrt{2})^2 = 4-2\sqrt{2}

and

f = 1+(1+\sqrt{2})^2 = 4+2\sqrt{2}

. For the last two points,

f = 1+0^2=1

. Comparing these values, the maximum is

4+2\sqrt{2}

Question 14

Let $f(x,y) = x^3 - 3x - y^2$ . Find the maximum value of $f$ on the line segment defined by $x=2$ and $-2 \le y \le 2$ .

$2$ (correct answer)
$1$
$-2$
$-3$

Explanation: This is a constrained optimization problem, but it can be solved by parameterization. The constraint is the line segment where

x=2

and

y

varies from

-2

2

. We substitute

x=2

into the function

f(x,y)

to get a function of a single variable,

y

h(y) = f(2,y) = 2^3 - 3(2) - y^2 = 8 - 6 - y^2 = 2 - y^2

. We need to find the maximum value of

h(y) = 2 - y^2

on the interval

[-2, 2]

. The derivative is

h'(y) = -2y

. Setting

h'(y)=0

gives

y=0

, which is a critical point within the interval. We evaluate

h(y)

at this critical point and at the endpoints of the interval:

h(0) = 2 - 0^2 = 2

h(-2) = 2 - (-2)^2 = 2 - 4 = -2

h(2) = 2 - 2^2 = 2 - 4 = -2

. The maximum value among these is 2. A student might try to find unconstrained critical points of

f(x,y)

\nabla f = \langle 3x^2-3, -2y \rangle = \langle 0,0 \rangle

, which gives

(1,0)

and

(-1,0)

. These points are not on the constraint line

x=2

Question 15

Let $Q$ be the point on the surface of the paraboloid $z = x^2 + y^2$ that is closest to the point $P(0, 0, 5)$ . Which of the following statements correctly describes the vector $\vec{PQ}$ ?

The vector $\vec{PQ}$ is parallel to the normal vector of the paraboloid at point $Q$ . (correct answer)
The vector $\vec{PQ}$ is perpendicular to the normal vector of the paraboloid at point $Q$ .
The vector $\vec{PQ}$ is parallel to the $xy$ -plane.
The point $Q$ is the vertex of the paraboloid, $(0, 0, 0)$ .

Explanation: To find the point

Q(x, y, z)

on the paraboloid closest to

P(0, 0, 5)

, we minimize the square of the distance function

f(x, y, z) = x^2 + y^2 + (z-5)^2

subject to the constraint

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

. Using the method of Lagrange multipliers, we set

\nabla f = \lambda \nabla g

. The gradient of

f

\nabla f = \langle 2x, 2y, 2(z-5) \rangle

, which is twice the vector

\vec{QP} = \langle x, y, z-5 \rangle = -\vec{PQ}

. The gradient of

g

\nabla g = \langle 2x, 2y, -1 \rangle

, which is a normal vector to the surface at

(x, y, z)

. The condition

\nabla f = \lambda \nabla g

implies that

\vec{QP}

is parallel to

\nabla g

. Therefore, the vector

\vec{PQ}

is parallel to the normal vector of the paraboloid at point

Q

. Explicitly solving, we find that the minimum distance occurs for all points on the circle

x^2+y^2=4.5

at height

z=4.5

, not at the vertex

(0,0,0)