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Calculus 3 : Lagrange Multipliers

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Example Questions

Example Question #3 : Applications Of Partial Derivatives

Find the minimum and maximum of f(x,y)=2x-5y , subject to the constraint x^2+y^2=144 .

Possible Answers:

f(4.46,11.14) is a maximum

f(-4.46,-11.14) is a minimum

There are no maximums or minimums

f(-4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

f(4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

Correct answer:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

Explanation:

First we need to set up our system of equations.

$2=2\lambda x \rightarrow x=\frac{1}{\lambda}$

$-5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}$

x^2+y^2=144

Now lets plug in these constraints.

$(\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144$

$\frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144$

$\frac{29}{4\lambda^2}=144$

Now we solve for $\lambda$

$\frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}$

$\lambda=\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14$

$\lambda=-\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14$

Now lets plug in these values of , and into the original equation.

f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62

f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62

We can conclude from this that f(4.46,-11.14) is a maximum, and f(-4.46,11.14) is a minimum.

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Example Question #1 : Lagrange Multipliers

Find the absolute minimum value of the function f(x,y) = x^2+y^2 subject to the constraint x^2 +2y^2 = 1 .

Possible Answers:

$\sqrt2$

$2\sqrt2$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

Let g = x^2 +2y^2 To find the absolute minimum value, we must solve the system of equations given by

$\triangledown f = \lambda\triangledown g, g =1$ .

So this system of equations is

$f_x = \lambda g_x$ , $f_y = \lambda g_y$ , g =1 .

Taking partial derivatives and substituting as indicated, this becomes

$2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1$ .

From the left equation, we see either x=0 or $\lambda = 1$ . If x=0 , then substituting this into the other equations, we can solve for $y, \lambda$ , and get $y = \pm \sqrt{2}/2$ , $\lambda = 1/2$ , giving two extreme candidate points at $(0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})$ .

On the other hand, if instead $\lambda =1$ , this forces y = 0 from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; (-1,0),(1,0) .

Taking all four of our found points, and plugging them back into , we have

$f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}$ .

Hence the absolute minimum value is $\frac{1}{2}$ .

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Example Question #1 : Lagrange Multipliers

Find the dimensions of a box with maximum volume such that the sum of its edges is cm.

Possible Answers:

$6 \times 6 \times 6$

$5 \times 4 \times 3$

$5 \times 5 \times 5$

$6 \times 5 \times 4$

Correct answer:

$5 \times 5 \times 5$

Explanation:

Untitled

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Example Question #2 : Lagrange Multipliers

Optimize $f(x)=\sqrt{6-x^2-y^2}$ using the constraint x+y-2=0

Possible Answers:

$\pm2$

$\pm\frac{1}{2}$

Correct answer:

$\pm2$

Explanation:

Untitled

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Example Question #3 : Lagrange Multipliers

Maximize f(x,y,z)=xyz with constraint x^2+2y^2+3z^2=6

Possible Answers:

$\frac{3}{2\sqrt{2}}$

$3\sqrt{2}$

$\frac{3}{2}$

$\frac{\sqrt{3}}{2}$

Correct answer:

$\frac{3}{2\sqrt{2}}$

Explanation:

Untitled

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Example Question #4 : Lagrange Multipliers

A company has the production function $P=100L^{\frac{3}{4}}K^{\frac{1}{4}}$ , where represents the number of hours of labor, and represents the capital. Each labor hour costs $150 and each unit capital costs $250. If the total cost of labor and capital is is $50,000, then find the maximum production.

Possible Answers:

$\$12500$

none of the above

$\$12300$

$\$12000$

Correct answer:

$\$12300$

Explanation:

Untitled

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Example Question #14 : Applications Of Partial Derivatives

Find the maximum value of the function f=x^2y with the constraint x^2+y^2=4 .

Possible Answers:

x=1 , $y=\sqrt{3}$

x=2 , y=2

$x=\frac{2\sqrt{6}}{3}$ , $y=\frac{2\sqrt{3}}{3}$

$x=\sqrt{3}$ , y=1

Correct answer:

$x=\frac{2\sqrt{6}}{3}$ , $y=\frac{2\sqrt{3}}{3}$

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_x=\lambda g_x$ and $f_y=\lambda g_y$ .

The equation being optimized is f=x^2y .

The constraint is x^2+y^2=4 .

f_x=2xy , f_y=x^2 , g_x=2x , g_y=2y

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$2xy=\lambda (2x)$

$x^2=\lambda (2y)$

x^2+y^2=4

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

$2xy=\lambda (2x)\Rightarrow y=\lambda$

$x^2=\lambda (2y)\Rightarrow x^2/2y=\lambda$

Setting the two expressions for $\lambda$ equal to each other gives us

y=x^2/2y

2y^2=x^2

Substituting this expression into the constraint gives us

x^2+y^2=4

2y^2+y^2=4

$3y^2=4\Rightarrow y^2=4/3$

$y=2/\sqrt{3}=\frac{2\sqrt{3}}{3}$

x^2=2y^2=2(4/3)=8/3

$x=\sqrt{\frac{8}{3}}=\frac{2\sqrt{2}}{\sqrt{3}}=\frac{2\sqrt{6}}{3}$

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Example Question #2 : Lagrange Multipliers

Find the maximum value of the function f=x^2+y^2-xy with the constraint 3x+4y=12 .

Possible Answers:

$x=\frac{60}{37}$ , $y=\frac{66}{37}$

$x=\frac{110}{37}$ , $y=\frac{121}{37}$

$x=\frac{8}{3}$ , y=1

x=2 , y=1.5

Correct answer:

$x=\frac{60}{37}$ , $y=\frac{66}{37}$

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_x=\lambda g_x$ and $f_y=\lambda g_y$ .

The equation being optimized is f=x^2+y^2-xy .

The constraint is 3x+4y=12 .

f_x=2x-y , f_y=2y-x , g_x=3 , g_y=4

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$2x-y=\lambda (3)$

$2y-x=\lambda (4)$

3x+4y=12

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

$2x-y=\lambda (3)\Rightarrow 2x/3-y/3=\lambda$

$2y-x=\lambda (4) \Rightarrow y/2-x/4=\lambda$

Setting the two expressions for $\lambda$ equal to each other gives us

2x/3-y/3=y/2-x/4

2x/3+x/4=y/2+y/3

$\frac{11x}{12}=\frac{5y}{6}$

$x=\frac{10y}{11}$

Substituting this expression into the constraint gives us

3x+4y=12

3(10y/11)+4y=12

30y/11+4y=12

74y/11=12

$y=12*11/74=\frac{66}{37}$

$x=\frac{10}{11}*\frac{66}{37}=\frac{60}{37}$

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Example Question #16 : Applications Of Partial Derivatives

A company makes chairs () and benches (). The profit equation for this company is f=-2c^2-3b^2+24c+28b+11 . The company can only produce pieces per day. How many of each seat should the company produce to maximize profit?

Possible Answers:

b= 14 , c= 20

b= 17 , c= 18

b= 8 , c= 27

b= 25 , c= 9

Correct answer:

b= 14 , c= 20

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_x=\lambda g_x$ and $f_y=\lambda g_y$ .

In this problem, we are trying to maximize the profit, so the equation being optimized is f=-2c^2-3b^2+24c+28b+11 .

The company can only produce pieces of furniture, so the constraint is c+b=35 .

f_c=-4c+24

f_b=-6b+28

g_c=1 , g_b=1

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$-4c+24=\lambda (1)$

$-6b+28=\lambda (1)$

c+b=35

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

Setting the two expressions of $\lambda$ equal to each other gives us

-4c+24=-6b+28

6b-28=4c-24

6b-4=4c

c=6b/4-1

Substituting this expression into the constraint gives

c+b=35

6b/4-1+b=35

10b/4=36

$b=14.4 \approx 14$

$c=6(14.4)/4-1=20.6 \approx 20$

Profit is maximized by making chairs and benches.

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Example Question #17 : Applications Of Partial Derivatives

A company makes end tables () and side tables (). The profit equation for this company is f=-7e^2-2s^2+10e+15s+55 . The company can only produce pieces per day. How many of each table should the company produce to maximize profit?

Possible Answers:

e=25 , s=14

e=4 , s=35

e=15 , s=24

e=10 , s=39

Correct answer:

e=10 , s=39

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_x=\lambda g_x$ and $f_y=\lambda g_y$ .

In this problem, we are trying to maximize the profit, so the equation being optimized is f=-7e^2-2s^2+10e+15s+55 .

The company can only produce pieces of furniture, so the constraint is e+s=50 .

f_e=-14e+10

f_s=-4s+15

g_e=1 , g_s=1

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$-14e+10=\lambda(1)$

$-4s+15=\lambda (1)$

e+s=50

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

Setting the two expressions of $\lambda$ equal to each other gives us

-14e+10=-4s+15

4s-15=14e-10

4s=14e+5

s=7e/2+5/4

Substituting this expression into the constraint gives

e+s=50

e+7e/2+5/4=50

9e/2+5/4=50

9e/2=195/4

$e=65/6 \approx 10$

$s=\frac{7}{2}*\frac{65}{6}+\frac{5}{4}=\frac{455}{12}+\frac{5}{4}=\frac{470}{12} \approx 39$

Profit is maximized by making end tables and side tables.

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Calculus 3 : Lagrange Multipliers

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3 : Applications Of Partial Derivatives

Example Question #1 : Lagrange Multipliers

Example Question #1 : Lagrange Multipliers

Example Question #2 : Lagrange Multipliers

Example Question #3 : Lagrange Multipliers

Example Question #4 : Lagrange Multipliers

Example Question #14 : Applications Of Partial Derivatives

Example Question #2 : Lagrange Multipliers

Example Question #16 : Applications Of Partial Derivatives

Example Question #17 : Applications Of Partial Derivatives

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