Lagrange Multipliers - Calculus 3

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Question

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

Answer

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 $\lambda$

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

If

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

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

If

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

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

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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Question

Find the absolute minimum value of the function subject to the constraint .

Answer

Let 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

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

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 or $\lambda = 1$ . If , 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 from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; .

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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Question

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

Answer

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Question

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

Answer

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Question

Maximize with constraint

Answer

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Question

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.

Answer

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Question

What is the least amount of wood required to make a rectangular sandbox whose area is ?

Answer

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 minimize the perimeter of the sandbox, so the equation being optimized is .

The constraint is the area of the box, or .

, , ,

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

$2=\lambda y$

$2=\lambda x$

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

$2=\lambda y\Rightarrow \lambda=2/y$

$2=\lambda x\Rightarrow \lambda=2/x$

$\frac{2}{y}=\frac{2}{x}\Rightarrow x=y$

$xy=16 \Rightarrow y^2=16$

These dimensions minimize the perimeter of the sandbox.

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Question

What is the least amount of fence required to make a yard bordered on one side by a house? The area of the yard is .

Answer

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 minimize the perimeter of the yard, which is three sides, so the equation being optimized is .

The constraint is the area of the fence, or .

, , ,

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

$2=\lambda y$

$1=\lambda x$

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

$2=\lambda y\Rightarrow \lambda=2/y$

$1=\lambda x\Rightarrow \lambda=1/x$

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

$\frac{2}{y}=\frac{1}{x}\Rightarrow x=y/2$

Substituting this expression into the constraint gives us

$xy=72 \Rightarrow \frac{1}{2}y^2=72$

These dimensions minimize the perimeter of the yard.

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Question

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

Answer

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.

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Question

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?

Answer

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.

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Question

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

Answer

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.

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Question

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?

Answer

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.

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Question

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.

Answer

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.

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Question

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.

Answer

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.

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Question

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?

Answer

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.

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Question

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

Answer

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

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

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

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

$b=14.4 \approx 14$

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

Profit is maximized by making chairs and benches.

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Question

Find the maximum value of the function with the constraint .

Answer

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

$y=1211/74=\frac{66}{37}$

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

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Question

Find the maximum value of the function with the constraint .

Answer

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

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

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

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

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