# Calculus 3 : Lagrange Multipliers

## Example Questions

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### Example Question #18 : Applications Of Partial Derivatives

Production is modeled by the function,  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, , where  is the Lagrangian multiplier.

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

and .

In this problem, we are trying to maximize the production, so the equation being optimized is .

We have a finite amount of money to purchase labor and capital, so the constraint is  .

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

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

Solving the first two equations for lambda gives

Setting the two expressions of  equal to each other gives us

Substituting this expression into the constraint gives

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

### Example Question #19 : Applications Of Partial Derivatives

Production is modeled by the function  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, , where  is the Lagrangian multiplier.

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

and .

In this problem, we are trying to maximize the production, so the equation being optimized is .

We have a finite amount of money to purchase labor and capital, so the constraint is  .

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

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

Solving the first two equations for lambda gives

Substituting this expression into the constraint gives

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

### Example Question #20 : Applications Of Partial Derivatives

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, , where  is the Lagrangian multiplier.

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

and .

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

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

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

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

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

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

Substituting  and  into the constraint gives us

These dimensions maximize the volume of the box.

### Example Question #21 : Applications Of Partial Derivatives

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, , where  is the Lagrangian multiplier.

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

and .

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

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

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

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

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

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

We now know . Substituting  and  into the constraint gives us

These dimensions maximize the volume of the box.

### Example Question #22 : Applications Of Partial Derivatives

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, , where  is the Lagrangian multiplier.

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

and .

In this problem, we are trying to minimize the surface area of the fish tank with no top, so the equation being optimized is .

The constraint is the volume of the cylinder, or .

,

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

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

Setting both expressions of lambda equal to each other gives us

Substituting this expression into the constraint, we have

These dimensions minimize the surface area of the fish tank.

### Example Question #23 : Applications Of Partial Derivatives

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, , where  is the Lagrangian multiplier.

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

and .

In this problem, we are trying to minimize the surface area of the soda can, so the equation being optimized is .

The constraint is the volume of the cylinder, or .

,

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

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

Setting both expressions of lambda equal to each other gives us

Substituting this expression into the constraint, we have

These dimensions minimize the surface area of the soda can.

### Example Question #24 : Applications Of Partial Derivatives

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 .

Explanation:

To optimize a function  subject to the constraint , we use the Lagrangian function, , where  is the Lagrangian multiplier.

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

and .

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

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

Setting the two expressions for  equal to each other gives us

Substituting this expression into the constraint gives us

These dimensions minimize the perimeter of the yard.

### Example Question #25 : Applications Of Partial Derivatives

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

Explanation:

To optimize a function  subject to the constraint , we use the Lagrangian function, , where  is the Lagrangian multiplier.

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

and .

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

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

These dimensions minimize the perimeter of the sandbox.

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