### All Calculus 3 Resources

## Example Questions

### Example Question #11 : Applications Of Partial Derivatives

Optimize using the constraint

**Possible Answers:**

**Correct answer:**

### Example Question #12 : Applications Of Partial Derivatives

Maximize with constraint

**Possible Answers:**

**Correct answer:**

### Example Question #13 : Applications Of Partial Derivatives

A company has the production function , 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

**Correct answer:**

### Example Question #14 : Applications Of Partial Derivatives

Find the maximum value of the function with the constraint .

**Possible Answers:**

,

,

,

,

**Correct answer:**

,

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 .

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

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

### Example Question #15 : Applications Of Partial Derivatives

Find the maximum value of the function with the constraint .

**Possible Answers:**

,

,

,

,

**Correct answer:**

,

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 .

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

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

### Example Question #16 : Applications Of Partial Derivatives

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?

**Possible Answers:**

,

,

,

,

**Correct answer:**

,

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 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

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

Setting the two expressions of equal to each other gives us

Substituting this expression into the constraint gives

Profit is maximized by making chairs and benches.

### Example Question #17 : Applications Of Partial Derivatives

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?

**Possible Answers:**

,

,

,

,

**Correct answer:**

,

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

and .

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 .

,

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

Setting the two expressions of equal to each other gives us

Substituting this expression into the constraint gives

Profit is maximized by making end tables and side tables.

### 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.

**Possible Answers:**

,

,

,

,

**Correct answer:**

,

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 .

,

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.

**Possible Answers:**

,

,

,

,

**Correct answer:**

,

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 .

,

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?

**Possible Answers:**

, ,

, ,

, ,

, ,

**Correct answer:**

, ,

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 .

, , ,

, ,

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.