Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #11 : Applications Of Partial Derivatives

Optimize  using the constraint 

Possible Answers:

Correct answer:

Explanation:

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Example Question #12 : Applications Of Partial Derivatives

Maximize  with constraint 

Possible Answers:

Correct answer:

Explanation:

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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:

 

Explanation:

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Example Question #14 : Applications Of Partial Derivatives

Find the maximum value of the function  with the constraint .

Possible Answers:

Correct answer:

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 .

 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:

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 .

 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:

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 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:

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 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  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:

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.

Possible Answers:

Correct answer:

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?

Possible Answers:

Correct answer:

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.

 

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