All Calculus 3 Resources
Example Questions
Example Question #7 :Applications Of Partial Derivatives
Find the minimum and maximum of, subject to the constraint.
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
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
如果
,
如果
,
Now lets plug in these values of, andinto the original equation.
We can conclude from this thatis a maximum, andis a minimum.
Example Question #8 :Applications Of Partial Derivatives
Find the absolute minimum value of the functionsubject to the constraint.
LetTo 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 eitheror. 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 forcesfrom the 2nd equation, andfrom 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 iscm.
Example Question #11 :Applications Of Partial Derivatives
Optimizeusing the constraint
Example Question #12 :Applications Of Partial Derivatives
Maximizewith constraint
Example Question #13 :Applications Of Partial Derivatives
A company has the production function, whererepresents the number of hours of labor, andrepresents 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.
none of the above
Example Question #14 :Applications Of Partial Derivatives
Find the maximum value of the functionwith the constraint.
,
,
,
,
,
To optimize a functionsubject to the constraint, we use the Lagrangian function,, whereis the Lagrangian multiplier.
如果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 forequal 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 functionwith the constraint.
,
,
,
,
,
To optimize a functionsubject to the constraint, we use the Lagrangian function,, whereis the Lagrangian multiplier.
如果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 forequal 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 producepieces per day. How many of each seat should the company produce to maximize profit?
,
,
,
,
,
To optimize a functionsubject to the constraint, we use the Lagrangian function,, whereis the Lagrangian multiplier.
如果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 producepieces 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 ofequal to each other gives us
Substituting this expression into the constraint gives
Profit is maximized by makingchairs andbenches.
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 producepieces per day. How many of each table should the company produce to maximize profit?
,
,
,
,
,
To optimize a functionsubject to the constraint, we use the Lagrangian function,, whereis the Lagrangian multiplier.
如果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 producepieces 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 ofequal to each other gives us
Substituting this expression into the constraint gives
Profit is maximized by makingend tables and副表。
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