Lagrange Multipliers

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AP Calculus BC › Lagrange Multipliers

Questions 1 - 10

A soda can (a right cylinder) has a volume of . What height and radius will minimize the surface area of the soda can?

Explanation

To optimize a function subject to the constraint , 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,

$f_r=\lambda g_r$ and $f_h=\lambda g_h$ .

In this problem, we are trying to minimize the surface area of the soda can, so the equation being optimized is $f=2\pi rh+2\pi r^2$ .

The constraint is the volume of the cylinder, or $\pi r^2h=0.5$ .

$f_r=2\pi h+4\pi r$ , $f_h=2\pi r$ , $g_r=2\pi rh$ , $g_h=\pi r^2$

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

$2\pi h+4\pi r=\lambda\left ( 2\pi rh\right )$

$2\pi r=\lambda \left (\pi r^2 \right )$

$\pi r^2h=0.5$

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

$2\pi h+4\pi r=\lambda\left ( 2\pi rh\right )\Rightarrow \frac{2\pi h+4\pi r}{ 2\pi rh}=\lambda$

$\frac{h+2r}{ rh}=\lambda$

$2\pi r=\lambda \left (\pi r^2 \right )\Rightarrow \frac{2\pi r}{\pi r^2}= \lambda$

$\frac{2}{ r}= \lambda$

Setting both expressions of lambda equal to each other gives us

$\frac{h+2r}{ rh}=\frac{2}{r}\Rightarrow r(h+2r)=2rh\Rightarrow rh+2r^2=2rh$

Substituting this expression into the constraint, we have

$\pi r^2h=0.5\Rightarrow \pi r^2(2r)=0.5$

$2\pi r^3=0.5\Rightarrow r=0.43dm$

These dimensions minimize the surface area of the soda can.

A fish tank (right cylinder) with no top has a volume of . What height and radius will minimize the surface area of the fish tank?

Explanation

To optimize a function subject to the constraint , 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,

$f_r=\lambda g_r$ and $f_h=\lambda g_h$ .

In this problem, we are trying to minimize the surface area of the fish tank with no top, so the equation being optimized is $f=2\pi rh+\pi r^2$ .

The constraint is the volume of the cylinder, or $\pi r^2h=27$ .

$f_r=2\pi h+2\pi r$ , $f_h=2\pi r$ , $g_r=2\pi rh$ , $g_h=\pi r^2$

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

$2\pi h+2\pi r=\lambda\left ( 2\pi rh\right )$

$2\pi r=\lambda \left (\pi r^2 \right )$

$\pi r^2h=27$

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

$2\pi h+2\pi r=\lambda\left ( 2\pi rh\right )\Rightarrow \frac{2\pi h+2\pi r}{ 2\pi rh}=\lambda$

$\frac{h+r}{ rh}=\lambda$

$2\pi r=\lambda \left (\pi r^2 \right )\Rightarrow \frac{2\pi r}{\pi r^2}= \lambda$

$\frac{2}{ r}= \lambda$

Setting both expressions of lambda equal to each other gives us

$\frac{h+r}{ rh}=\frac{2}{r}\Rightarrow r(h+r)=2rh\Rightarrow rh+r^2=2rh$

Substituting this expression into the constraint, we have

$\pi r^2h=27\Rightarrow \pi r^2(r)=27$

$\pi r^3=27\Rightarrow r=2.05m$

These dimensions minimize the surface area of the fish tank.

A box has a surface area of . What length, width and height maximize the volume of the box?

, ,

Explanation

To optimize a function subject to the constraint , 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 box, so the equation being optimized is .

The constraint is the surface area of the box, or .

, , ,

, ,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We now know . Substituting and into the constraint gives us

$x^2=20.833\Rightarrow x=4.56in$

These dimensions maximize the volume of the box.

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 and each unit of capital costs . If a company has to spend, how many units of labor and capital should be purchased.

Explanation

To optimize a function subject to the constraint , 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 .

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

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

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

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

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

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

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

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

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

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

Substituting this expression into the constraint gives

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

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

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

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 and each unit of capital costs . If a company has to spend, how many units of labor and capital should be purchased.

Explanation

To optimize a function subject to the constraint , 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 .

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

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

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

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

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

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

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

$l=160/3 \approx 53$

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

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

A company makes end tables () and side tables (). The profit equation for this company is . The company can only produce pieces per day. How many of each table should the company produce to maximize profit?

Explanation

To optimize a function subject to the constraint , 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 .

The company can only produce pieces of furniture, so the constraint is .

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

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

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

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

Substituting this expression into the constraint gives

$e=65/6 \approx 10$

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

Profit is maximized by making end tables and side tables.

Find the maximum value of the function with the constraint .

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

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

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

Explanation

To optimize a function subject to the constraint , 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 .

The constraint is .

, , ,

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

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

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

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

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

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

Substituting this expression into the constraint gives us

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

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

Find the maximum value of the function with the constraint .

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

, $y=\sqrt{3}$

$x=\sqrt{3}$ ,

Explanation

To optimize a function subject to the constraint , 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 .

The constraint is .

, , ,

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

$2xy=\lambda (2x)$

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

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

Substituting this expression into the constraint gives us

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

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

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

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

, ,

Explanation

To optimize a function subject to the constraint , 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 .

The constraint is the surface area of the box with no bottom, or .

, , ,

, ,

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

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

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

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

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

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

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

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

Substituting and into the constraint gives us

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

These dimensions maximize the volume of the box.

Find the minimum and maximum of , subject to the constraint .

is a maximum

is a minimum

is a maximum

is a minimum

There are no maximums or minimums

is a maximum

is a minimum

is a maximum

is a minimum

Explanation

First we need to set up our system of equations.

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

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

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.

We can conclude from this that is a maximum, and is a minimum.

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