### All Calculus 3 Resources

## Example Questions

### Example Question #7 : Applications Of Partial Derivatives

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

**Possible Answers:**

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

**Correct answer:**

is a maximum

is a minimum

First we need to set up our system of equations.

Now lets plug in these constraints.

Now we solve for

If

,

If

,

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.

### Example Question #8 : Applications Of Partial Derivatives

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

**Possible Answers:**

**Correct answer:**

Let To find the absolute minimum value, we must solve the system of equations given by

.

So this system of equations is

, , .

Taking partial derivatives and substituting as indicated, this becomes

.

From the left equation, we see either or . If , then substituting this into the other equations, we can solve for , and get , , giving two extreme candidate points at .

On the other hand, if instead , this forces from the 2nd equation, and 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

.

Hence the absolute minimum value is .

### Example Question #9 : Applications Of Partial Derivatives

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

**Possible Answers:**

**Correct answer:**

### 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.