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Calculus 3 : Applications of Partial Derivatives

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Example Questions

Example Question #1 : Lagrange Multipliers

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

Possible Answers:

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

$\pm2$

Correct answer:

$\pm2$

Explanation:

Untitled

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Example Question #2 : Lagrange Multipliers

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

Possible Answers:

$3\sqrt{2}$

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

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

$\frac{3}{2}$

Correct answer:

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

Explanation:

Untitled

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Example Question #11 : Applications Of Partial Derivatives

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:

none of the above

$\$12300$

$\$12000$

$\$12500$

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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Example Question #18 : Applications Of Partial Derivatives

Production is modeled by the function, $f=50l^{1/5}c^{4/5}$ where is the units of labor and is the units of capital. Each unit of labor costs $\$15$ and each unit of capital costs $\$18$ . If a company has $\$4,000$ to spend, how many units of labor and capital should be purchased.

Possible Answers:

l=47 , c= 171

l=53 , c= 177

l=155 , c= 75

l=65 , c= 165

Correct answer:

l=53 , c= 177

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 production, so the equation being optimized is $f=50l^{1/5}c^{4/5}$ .

We have a finite amount of money to purchase labor and capital, so the constraint is 15l+18c=4000 .

$f_l=(1/5)(50)l^{-4/5}c^{4/5}=10\left (\frac{c}{l} \right )^{4/5}$

$f_c=(4/5)(50)l^{1/5}c^{-1/5}=40\left (\frac{c}{l} \right )^{-1/5}$

g_l=15 , g_c=18

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

$10\left (\frac{c}{l} \right )^{4/5}=\lambda (15)$

$40\left (\frac{c}{l} \right )^{-1/5}=\lambda (18)$

15l+18c=4000

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

Solving the first two equations for lambda gives

$10\left (\frac{c}{l} \right )^{4/5}=\lambda (15)\Rightarrow \left ( \frac{2}{3} \right )\left (\frac{c}{l} \right )^{4/5}= \lambda$

$40\left (\frac{c}{l} \right )^{-1/5}=\lambda (18)\Rightarrow \left (\frac{20}{9} \right )\left (\frac{c}{l} \right )^{-1/5}=\lambda$

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

$\left ( \frac{2}{3} \right )\left (\frac{c}{l} \right )^{4/5}= \left (\frac{20}{9} \right )\left (\frac{c}{l} \right )^{-1/5}$

$\left (\frac{c}{l} \right )= \left (\frac{20}{9} \right )\left ( \frac{3}{2} \right )=\left ( \frac{10}{3} \right )$

$c=\left ( \frac{10}{3} \right )l$

Substituting this expression into the constraint gives

15l+18c=4000

$15l+18\left ( \frac{10}{3} \right )l=4000$

15l+60l=4000

75l=4000

$l=160/3 \approx 53$

$c=\left ( \frac{10}{3} \right )\frac{160}{3}=\frac{1600}{9} \approx 177$

Buying units of labor and 177 units of capital will maximize production.

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Example Question #11 : Applications Of Partial Derivatives

Production is modeled by the function $f=75l^{3/4}c^{1/4}$ where is the units of labor and is the units of capital. Each unit of labor costs $\$100$ and each unit of capital costs $\$125$ . If a company has $\$20,000$ to spend, how many units of labor and capital should be purchased.

Possible Answers:

l=59 , c = 122

l=109 , c = 72

l=218 , c = 36

l=128 , c = 53

Correct answer:

l=109 , c = 72

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 production, so the equation being optimized is $f=75l^{3/4}c^{1/4}$ .

We have a finite amount of money to purchase labor and capital, so the constraint is 100l+125c=20000 .

$f_l=(3/4)(75)l^{-1/4}c^{1/4}=56.25\frac{c^{1/4}}{l^{1/4}}$

$f_c=(1/4)(75)l^{3/4}c^{-3/4}=18.75\frac{l^{3/4}}{c^{3/4}}$

g_l=100 , g_c=125

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

$56.25\frac{c^{1/4}}{l^{1/4}}=\lambda(100)$

$18.75\frac{l^{3/4}}{c^{3/4}}=\lambda(125)$

100l+125c=20000

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

Solving the first two equations for lambda gives

$56.25\frac{c^{1/4}}{l^{1/4}}=\lambda(100)\Rightarrow 0. 5625\left (\frac{c}{l} \right )^{1/4}=\lambda$

$18.75\frac{l^{3/4}}{c^{3/4}}=\lambda(125)\Rightarrow 0.15\left (\frac{l}{c} \right )^{3/4}=\lambda\Rightarrow 0.15\left (\frac{c}{l} \right )^{-3/4}=\lambda$

$0. 5625\left (\frac{c}{l} \right )^{1/4}=0.15\left (\frac{c}{l} \right )^{-3/4}$

$\frac{c}{l} =0.15/0.5625$

c =2l/3

Substituting this expression into the constraint gives

100l+125c=20000

100l+125(2l/3)=20000

100l+250l/3=20000

$550l/3=20000\Rightarrow l=60000/550=1200/11 \approx \109$

$c =2l/3\Rightarrow c=\frac{2}{3}\left (\frac{1200}{11} \right )=\frac{800}{11} \approx 72$

Buying 109 units of labor and units of capital will maximize production.

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Example Question #20 : Applications Of Partial Derivatives

A tiger cage is being built at the zoo (it has no bottom). Its surface area is 1728m^3 . What dimensions maximize the surface area of the box?

Possible Answers:

x=19.32m , y=24.74m , z=9.31m

x=34.12m , y=11.36m , z=15.8m

x=20.78m , y=20.78m , z=10.39m

x=47.31m , y=21.34m , z=8.18m

Correct answer:

x=20.78m , y=20.78m , z=10.39m

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 three-dimensional function, the Lagrangian function expands to three equations,

$f_x=\lambda g_x$ , $f_y=\lambda g_y$ and $f_z=\lambda g_z$ .

In this problem, we are trying to maximize the volume of the cage, so the equation being optimized is f=xyz .

The constraint is the surface area of the box with no bottom, or xy+2xz+2yz=1728 .

f_x=yz , f_y=xz , f_z=xy ,

g_x=y+2z , g_y=x+2z , g_z=2x+2y

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

$yz=\lambda (y+2z)$

$xz=\lambda (x+2z)$

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

2xy+2xz+2yz=1728

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

Multiplying the first equation by and the second equation by gives us

$xyz=\lambda x(y+2z)$

$xyz=\lambda y(x+2z)$

The left side of both equations are the same, so we can set the right sides equal to each other

$\lambda x(y+2z)=\lambda y(x+2z)$

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

xy+2xz=xy+2yz

2xz=2yz

x=y

Multiplying the first equation by and the second equation by gives us

$xyz=\lambda x(y+2z)$

$xyz=\lambda z(2x+2y)$

The left side of both equations are the same, so we can set the right sides equal to each other

$\lambda x(y+2z)=\lambda z(2x+2y)$

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

xy+2xz=2xz+2yz

xy=2yz

x=2z

Substituting y=x and z=x/2 into the constraint gives us

2xy+2xz+2yz=1728

2x(x)+2x(x/2)+2(x)(x/2)=1728

2x^2+x^2+x^2=1728

4x^2=1728

$x^2=432\Rightarrow x=20.78m$

y=20.78m

z=10.39m

These dimensions maximize the volume of the box.

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Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Lagrange Multipliers

Example Question #2 : Lagrange Multipliers

Example Question #11 : Applications Of Partial Derivatives

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

Example Question #18 : Applications Of Partial Derivatives

Example Question #11 : Applications Of Partial Derivatives

Example Question #20 : Applications Of Partial Derivatives

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