## Active Calculus - Multivariable

### Section10.8Constrained Optimization: Lagrange Multipliers

###### Motivating Questions

What geometric condition enables us to optimize a function \(f=f(x,y)\) subject to a constraint given by \(g(x,y) = k\text{,}\) where \(k\) is a constant?

How can we exploit this geometric condition to find the extreme values of a function subject to a constraint?

We previously considered how to find the extreme values of functions on both unrestricted domains and on closed, bounded domains. Other types of optimization problems involve maximizing or minimizing a quantity subject to an external constraint. In these cases the extreme values frequently won't occur at the points where the gradient is zero, but rather at other points that satisfy an important geometric condition. These problems are often called *constrained optimization* problems and can be solved with the method of Lagrange Multipliers, which we study in this section.

###### Preview Activity10.8.1.

According to U.S. postal regulations, the girth plus the length of a parcel sent by mail may not exceed 108 inches, where by “girth” we mean the perimeter of the smallest end. Our goal is to find the largest possible volume of a rectangular parcel with a square end that can be sent by mail. (We solved this applied optimization problem in single variable *Active Calculus*, so it may look familiar. We take a different approach in this section, and this approach allows us to view most applied optimization problems from single variable calculus as constrained optimization problems, as well as provide us tools to solve a greater variety of optimization problems.) If we let \(x\) be the length of the side of one square end of the package and \(y\) the length of the package, then we want to maximize the volume \(f(x,y) = x^2y\) of the box subject to the constraint that the girth (\(4x\)) plus the length (\(y\)) is as large as possible, or \(4x+y = 108\text{.}\) The equation \(4x + y = 108\) is thus an external constraint on the variables.

The constraint equation involves the function \(g\) that is given by

\begin{equation*} g(x,y) = 4x+y. \end{equation*}

Explain why the constraint is a contour of \(g\text{,}\) and is therefore a two-dimensional curve.

Figure 10.8.1 shows the graph of the constraint equation \(g(x,y) = 108\) along with a few contours of the volume function \(f\text{.}\) Since our goal is to find the maximum value of \(f\) subject to the constraint \(g(x,y) = 108\text{,}\) we want to find the point on our constraint curve that intersects the contours of \(f\) at which \(f\) has its largest value.

Points \(A\) and \(B\) in Figure 10.8.1 lie on a contour of \(f\) and on the constraint equation \(g(x,y) = 108\text{.}\) Explain why neither \(A\) nor \(B\) provides a maximum value of \(f\) that satisfies the constraint.

Points \(C\) and \(D\) in Figure 10.8.1 lie on a contour of \(f\) and on the constraint equation \(g(x,y) = 108\text{.}\) Explain why neither \(C\) nor \(D\) provides a maximum value of \(f\) that satisfies the constraint.

Based on your responses to parts i. and ii., draw the contour of \(f\) on which you believe \(f\) will achieve a maximum value subject to the constraint \(g(x,y) = 108\text{.}\) Explain why you drew the contour you did.

Recall that \(g(x,y) = 108\) is a contour of the function \(g\text{,}\) and that the gradient of a function is always orthogonal to its contours. With this in mind, how should \(\nabla f\) and \(\nabla g\) be related at the optimal point? Explain.

### Subsection10.8.1Constrained Optimization and Lagrange Multipliers

In Preview Activity 10.8.1, we considered an optimization problem where there is an external constraint on the variables, namely that the girth plus the length of the package cannot exceed 108 inches. We saw that we can create a function \(g\) from the constraint, specifically \(g(x,y) = 4x+y\text{.}\) The constraint equation is then just a contour of \(g\text{,}\) \(g(x, y) = c\text{,}\) where \(c\) is a constant (in our case 108). Figure 10.8.2 illustrates that the volume function \(f\) is maximized, subject to the constraint \(g(x, y) = c\text{,}\) when the graph of \(g(x, y) = c\) is tangent to a contour of \(f\text{.}\) Moreover, the value of \(f\) on this contour is the sought maximum value.

To find this point where the graph of the constraint is tangent to a contour of \(f\text{,}\) recall that \(\nabla f\) is perpendicular to the contours of \(f\) and \(\nabla g\) is perpendicular to the contour of \(g\text{.}\) At such a point, the vectors \(\nabla g\) and \(\nabla f\) are parallel, and thus we need to determine the points where this occurs. Recall that two vectors are parallel if one is a nonzero scalar multiple of the other, so we therefore look for values of a parameter \(\lambda\) that make

\begin{equation} \nabla f = \lambda \nabla g.\label{eq_10_8_Lagrange_ex1}\tag{10.8.1} \end{equation}

The constant \(\lambda\) is called a *Lagrange multiplier*.

To find the values of \(\lambda\) that satisfy (10.8.1) for the volume function in Preview Activity 10.8.1, we calculate both \(\nabla f\) and \(\nabla g\text{.}\) Observe that

\begin{equation*} \nabla f = 2xy \vi + x^2 \vj \ \ \ \ \text{ and } \ \ \ \ \nabla g = 4\vi + \vj, \end{equation*}

and thus we need a value of \(\lambda\) so that

\begin{equation*} 2xy \vi + x^2 \vj = \lambda(4\vi + \vj). \end{equation*}

Equating components in the most recent equation and incorporating the original constraint, we have three equations

\begin{align} 2xy \amp = \lambda (4) \label{eq_10_8_lag_ex1}\tag{10.8.2}\\ x^2 \amp = \lambda (1) \label{eq_10_8_lag_ex2}\tag{10.8.3}\\ 4x+y \amp = 108 \label{eq_10_8_lag_ex3}\tag{10.8.4} \end{align}

in the three unknowns \(x\text{,}\) \(y\text{,}\) and \(\lambda\text{.}\) First, note that if \(\lambda = 0\text{,}\) then equation (10.8.3) shows that \(x=0\text{.}\) From this, Equation (10.8.4) tells us that \(y = 108\text{.}\) So the point \((0,108)\) is a point we need to consider. Next, provided that \(\lambda \neq 0\) (from which it follows that \(x \neq 0\) by Equation (10.8.3)), we may divide both sides of Equation (10.8.2) by the corresponding sides of (10.8.3) to eliminate \(\lambda\text{,}\) and thus find that

\begin{align*} \frac{2y}{x} \amp = 4, \ \mbox{so}\\ y \amp = 2x. \end{align*}

Substituting into Equation (10.8.4) gives us

\begin{equation*} 4x+2x = 108 \end{equation*}

or

\begin{equation*} x = 18. \end{equation*}

Thus we have \(y = 2x = 36\) and \(\lambda = x^2 = 324\) as another point to consider. So the points at which the gradients of \(f\) and \(g\) are parallel, and thus at which \(f\) may have a maximum or minimum subject to the constraint, are \((0,108)\) and \((18,36)\text{.}\) By evaluating the function \(f\) at these points, we see that we maximize the volume when the length of the square end of the box is 18 inches and the length is 36 inches, for a maximum volume of \(f(18,36) = 11664\) cubic inches. Since \(f(0,108) = 0\text{,}\) we obtain a minimum value at this point.

We summarize the process of Lagrange multipliers as follows.

###### The method of Lagrange multipliers.

The general technique for optimizing a function \(f = f(x,y)\) subject to a constraint \(g(x,y)=c\) is to solve the system \(\nabla f = \lambda \nabla g\) and \(g(x,y)=c\) for \(x\text{,}\) \(y\text{,}\) and \(\lambda\text{.}\) We then evaluate the function \(f\) at each point \((x,y)\) that results from a solution to the system in order to find the optimum values of \(f\) subject to the constraint.

###### Activity10.8.2.

A cylindrical soda can holds about 355 cc of liquid. In this activity, we want to find the dimensions of such a can that will minimize the surface area. For the sake of simplicity, assume the can is a perfect cylinder.

What are the variables in this problem? Based on the context, what restriction(s), if any, are there on these variables?

What quantity do we want to optimize in this problem? What equation describes the constraint? (You need to decide which of these functions plays the role of \(f\) and which plays the role of \(g\) in our discussion of Lagrange multipliers.)

Find \(\lambda\) and the values of your variables that satisfy Equation (10.8.1) in the context of this problem.

Determine the dimensions of the pop can that give the desired solution to this constrained optimization problem.

The method of Lagrange multipliers also works for functions of more than two variables.

###### Activity10.8.3.

Use the method of Lagrange multipliers to find the dimensions of the least expensive packing crate with a volume of 240 cubic feet when the material for the top costs $2 per square foot, the bottom is $3 per square foot and the sides are $1.50 per square foot.

The method of Lagrange multipliers also works for functions of three variables. That is, if we have a function \(f = f(x,y,z)\) that we want to optimize subject to a constraint \(g(x,y,z) = k\text{,}\) the optimal point \((x,y,z)\) lies on the level surface \(S\) defined by the constraint \(g(x,y,z) = k\text{.}\) As we did in Preview Activity 10.8.1, we can argue that the optimal value occurs at the level surface \(f(x,y,z) = c\) that is tangent to \(S\text{.}\) Thus, the gradients of \(f\) and \(g\) are parallel at this optimal point. So, just as in the two variable case, we can optimize \(f = f(x,y,z)\) subject to the constraint \(g(x,y,z) = k\) by finding all points \((x,y,z)\) that satisfy \(\nabla f = \lambda \nabla g\) and \(g(x,y,z) = k\text{.}\)

### Subsection10.8.2Summary

The extrema of a function \(f=f(x,y)\) subject to a constraint \(g(x,y) = c\) occur at points for which the contour of \(f\) is tangent to the curve that represents the constraint equation. This occurs when

\begin{equation*} \nabla f = \lambda \nabla g. \end{equation*}

We use the condition \(\nabla f = \lambda \nabla g\) to generate a system of equations, together with the constraint \(g(x,y) = c\text{,}\) that may be solved for \(x\text{,}\) \(y\text{,}\) and \(\lambda\text{.}\) Once we have all the solutions, we evaluate \(f\) at each of the \((x,y)\) points to determine the extrema.

### Exercises10.8.3Exercises

###### 1.

Use Lagrange multipliers to find the maximum and minimum values of \(f(x,y) = 3 x - 4 y\) subject to the constraint \(x^2 + 3 y^2 = 129\text{,}\) if such values exist.

maximum =

minimum =

*(For either value, enter DNE if there is no such value.)*

###### 2.

Use Lagrange multipliers to find the maximum and minimum values of \(f(x,y) = x^2 y + 3 y^2 - y\text{,}\) subject to the constraint \(x^2 + y^2 \leq 38.3333333333333\)

maximum =

minimum =

*(For either value, enter DNE if there is no such value.)*

###### 3.

Find the absolute maximum and minimum of the function \(f(x,y) = x^2 + y^2\) subject to the constraint \(x^4 + y^4 = 6561\text{.}\)

As usual, ignore unneeded answer blanks, and list points in lexicographic order.

Absolute minimum value:

attained at (, ), (, ),

(, ), (, ).

Absolute maximum value:

attained at (, ), (, ),

(, ), (, ).

###### 4.

Find the absolute maximum and minimum of the function \(f(x,y) = x^2 - y^2\) subject to the constraint \(x^2 + y^2 = 361\text{.}\)

As usual, ignore unneeded answer blanks, and list points in lexicographic order.

Absolute minimum value:

attained at (, ) and (, ).

Absolute maximum value:

attained at (, ) and (, ).

###### 5.

Find the minimum distance from the point \((1,1,11)\) to the paraboloid given by the equation \(z=x^2+y^2\text{.}\)

Minimum distance =

*Note:* If you need to find roots of a polynomial of degree \(\geq 3\text{,}\) you may want to use a calculator of computer to do so numerically. Also be sure that you can give a geometric justification for your answer.

###### 6.

For each value of \(\lambda\) the function \(h(x,y) = x^2 + y^2 - \lambda(2 x + 8 y - 20)\) has a minimum value \(m(\lambda)\text{.}\)

*(a)* Find \(m(\lambda)\)

\(m(\lambda) =\)

*(Use the letter L for \(\lambda\) in your expression.)*

*(b)* For which value of \(\lambda\) is \(m(\lambda)\) the largest, and what is that maximum value?

\(\lambda =\)

maximum \(m(\lambda) =\)

*(c)* Find the minimum value of \(f(x,y)=x^2+y^2\) subject to the constraint \(2 x + 8 y = 20\) using the method of Lagrange multipliers and evaluate \(\lambda\text{.}\)

minimum \(f\) =

\(\lambda =\)

*(How are these results related to your result in part (b)?)*

\(20L-\frac{\left(2\cdot 2+8\cdot 8\right)L^{2}}{4}\)

\(\frac{2\cdot 20}{2\cdot 2+8\cdot 8}\)

\(\frac{20\cdot 20}{2\cdot 2+8\cdot 8}\)

\(\frac{20\cdot 20}{2\cdot 2+8\cdot 8}\)

\(\frac{2\cdot 20}{2\cdot 2+8\cdot 8}\)

###### 7.

The plane \(x + y + 2z = 6\) intersects the paraboloid \(z = x^2 + y^2\) in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

Point farthest away occurs at

(, ,).

Point nearest occurs at

(, ,).

###### 8.

Find the maximum and minimum values of the function \(f(x,y,z) = x^2 y^2 z^2\) subject to the constraint \(x^2 + y^2 + z^2 = 64\text{.}\)

Maximum value is , occuring at points (positive integer or "infinitely many").

Minimum value is , occuring at points (positive integer or "infinitely many").

###### 9.

Find the maximum and minimum values of the function \(f(x,y,z,t) = x+y+z+t\) subject to the constraint \(x^2 + y^2 + z^2 + t^2 = 100\text{.}\)

Maximum value is , occuring at points (positive integer or "infinitely many").

Minimum value is , occuring at points (positive integer or "infinitely many").

###### 10.

Find the maximum and minimum volumes of a rectangular box whose surface area equals 7000 square cm and whose edge length (sum of lengths of all edges) is 440 cm.

Hint: It can be deduced that the box is not a cube, so if x, y, and z are the lengths of the sides, you may want to let x represent a side with \(x \ne y\) and \(x \ne z\text{.}\)

Maximum value is ,

occuring at (, ,).

Minimum value is ,

occuring at (, ,).

###### 11.

*(a)* If \(\sum_{i=1}^3 x_i = 4\text{,}\) find the values of \(x_1, x_2, x_3\) making \(\sum_{i=1}^3 {x_i}^2\) minimum.

\(x_1, x_2, x_3 =\)

*(Give your values as a comma separated list.)*

*(b)* Generalize the result of part (a) to find the minimum value of \(\sum_{i=1}^n{x_i}^2\) subject to \(\sum_{i=1}^n x_i=4\text{.}\)

minimum value =

###### 12.

The Cobb-Douglas production function is used in economics to model production levels based on labor and equipment. Suppose we have a specific Cobb-Douglas function of the form

\begin{equation*} f(x, y) = 50 x^{0.4}y^{0.6}, \end{equation*}

where \(x\) is the dollar amount spent on labor and \(y\) the dollar amount spent on equipment. Use the method of Lagrange multipliers to determine how much should be spent on labor and how much on equipment to maximize productivity if we have a total of 1.5 million dollars to invest in labor and equipment.

###### 13.

Use the method of Lagrange multipliers to find the point on the line \(x-2y=5\) that is closest to the point \((1,3)\text{.}\) To do so, respond to the following prompts.

Write the function \(f=f(x,y)\) that measures the

*square*of the distance from \((x,y)\) to \((1,3)\text{.}\) (The extrema of this function are the same as the extrema of the distance function, but \(f(x,y)\) is simpler to work with.)What is the constraint \(g(x,y) = c\text{?}\)

Write the equations resulting from \(\nabla f = \lambda \nabla g\) and the constraint. Find all the points \((x,y)\) satisfying these equations.

Test all the points you found to determine the extrema.

###### 14.

Apply the Method of Lagrange Multipliers solve each of the following constrained optimization problems.

Determine the absolute maximum and absolute minimum values of \(f(x,y) = (x-1)^2 + (y-2)^2\) subject to the constraint that \(x^2 + y^2 = 16\text{.}\)

Determine the points on the sphere \(x^2 + y^2 + z^2 = 4\) that are closest to and farthest from the point \((3,1,-1)\text{.}\) (As in the preceding exercise, you may find it simpler to work with the square of the distance formula, rather than the distance formula itself.)

Find the absolute maximum and minimum of \(f(x,y,z) = x^2 + y^2 + z^2\) subject to the constraint that \((x-3)^2 + (y+2)^2 + (z-5)^2 \le 16\text{.}\) (Hint: here the constraint is a closed, bounded region. Use the boundary of that region for applying Lagrange Multipliers, but don't forget to also test any critical values of the function that lie in the interior of the region.)

###### 15.

In this exercise we consider how to apply the Method of Lagrange Multipliers to optimize functions of three variable subject to two constraints. Suppose we want to optimize \(f = f(x,y,z)\) subject to the constraints \(g(x,y,z) = c\) and \(h(x,y,z) = k\text{.}\) Also suppose that the two level surfaces \(g(x,y,z) = c\) and \(h(x,y,z) = k\) intersect at a curve \(C\text{.}\) The optimum point \(P = (x_0,y_0,z_0)\) will then lie on \(C\text{.}\)

Assume that \(C\) can be represented parametrically by a vector-valued function \(\vr = \vr(t)\text{.}\) Let \(\overrightarrow{OP} = \vr(t_0)\text{.}\) Use the Chain Rule applied to \(f(\vr(t))\text{,}\) \(g(\vr(t))\text{,}\) and \(h(\vr(t))\text{,}\) to explain why

\begin{align*} \nabla f(x_0,y_0,z_0) \cdot \vr'(t_0) \amp = 0, \\ \nabla g(x_0,y_0,z_0) \cdot \vr'(t_0) \amp = 0, \text{ and } \\ \nabla h(x_0,y_0,z_0) \cdot \vr'(t_0) \amp = 0. \end{align*}

Explain how this shows that \(\nabla f(x_0,y_0,z_0)\text{,}\) \(\nabla g(x_0,y_0,z_0)\text{,}\) and \(\nabla h(x_0,y_0,z_0)\) are all orthogonal to \(C\) at \(P\text{.}\) This shows that \(\nabla f(x_0,y_0,z_0)\text{,}\) \(\nabla g(x_0,y_0,z_0)\text{,}\) and \(\nabla h(x_0,y_0,z_0)\) all lie in the same plane.

Assuming that \(\nabla g(x_0,y_0,z_0)\) and \(\nabla h(x_0,y_0,z_0)\) are nonzero and not parallel, explain why every point in the plane determined by \(\nabla g(x_0,y_0,z_0)\) and \(\nabla h(x_0,y_0,z_0)\) has the form \(s\nabla g(x_0,y_0,z_0)+t\nabla h(x_0,y_0,z_0)\) for some scalars \(s\) and \(t\text{.}\)

Parts (a.) and (b.) show that there must exist scalars \(\lambda\) and \(\mu\) such that

\begin{equation*} \nabla f(x_0,y_0,z_0) = \lambda \nabla g(x_0,y_0,z_0)+ \mu \nabla h(x_0,y_0,z_0). \end{equation*}

So to optimize \(f = f(x,y,z)\) subject to the constraints \(g(x,y,z) = c\) and \(h(x,y,z) = k\) we must solve the system of equations

\begin{align*} \nabla f(x,y,z) \amp = \lambda \nabla g(x,y,z)+ \mu \nabla h(x,y,z), \\ g(x,y,z) \amp = c, \text{ and } \\ h(x,y,z) \amp = k. \end{align*}

for \(x\text{,}\) \(y\text{,}\) \(z\text{,}\) \(\lambda\text{,}\) and \(\mu\text{.}\)

Use this idea to find the maximum and minium values of \(f(x,y,z) = x+2y\) subject to the constraints \(y^2+z^2=8\) and \(x+y+z = 10\text{.}\)

###### 16.

There is a useful interpretation of the Lagrange multiplier \(\lambda\text{.}\) Assume that we want to optimize a function \(f\) with constraint \(g(x,y)=c\text{.}\) Recall that an optimal solution occurs at a point \((x_0, y_0)\) where \(\nabla f = \lambda \nabla g\text{.}\) As the constraint changes, so does the point at which the optimal solution occurs. So we can think of the optimal point as a function of the parameter \(c\text{,}\) that is \(x_0 = x_0(c)\) and \(y_0=y_0(c)\text{.}\) The optimal value of \(f\) subject to the constraint can then be considered as a function of \(c\) defined by \(f(x_0(c), y_0(c))\text{.}\) The Chain Rule shows that

\begin{equation*} \frac{df}{dc} = \frac{\partial f}{\partial x_0} \frac{dx_0}{dc} + \frac{\partial f}{\partial y_0} \frac{dy_0}{dc}. \end{equation*}

Use the fact that \(\nabla f = \lambda \nabla g\) at \((x_0,y_0)\) to explain why

\begin{equation*} \frac{df}{dc} = \lambda \frac{dg}{dc}. \end{equation*}

Use the fact that \(g(x,y) = c\) to show that

\begin{equation*} \frac{df}{dc} = \lambda. \end{equation*}

Conclude that \(\lambda\) tells us the rate of change of the function \(f\) as the parameter \(c\) increases (or by approximately how much the optimal value of the function \(f\) will change if we increase the value of \(c\) by 1 unit).

Suppose that \(\lambda = 324\) at the point where the package described in Preview Activity 10.8.1 has its maximum volume. Explain in context what the value \(324\) tells us about the package.

Suppose that the maximum value of a function \(f = f(x,y)\) subject to a constraint \(g(x,y) = 100\) is \(236\text{.}\) When using the method of Lagrange multipliers and solving \(\nabla f = \lambda \nabla g\text{,}\) we obtain a value of \(\lambda = 15\) at this maximum. Find an approximation to the maximum value of \(f\) subject to the constraint \(g(x,y) = 98\text{.}\)

## Introduction to Constrained Optimization in the Wolfram Language

### Optimization Problems

Constrained optimization problems are problems for which a function is to be minimized or maximized subject to constraints . Here is called the objective function and is a Boolean-valued formula. In the Wolfram Language the constraints can be an arbitrary Boolean combination of equations , weak inequalities , strict inequalities , and statements. The following notation will be used.

stands for "minimize subject to constraints ", and

stands for "maximize subject to constraints ".

You say a point satisfies the constraints if is true.

The following describes constrained optimization problems more precisely, restricting the discussion to minimization problems for brevity.

### Global Optimization

A point is said to be a global minimum of subject to constraints if satisfies the constraints and for any point that satisfies the constraints, .

A value is said to be the global minimum value of subject to constraints if for any point that satisfies the constraints, .

The global minimum value exists for any and . The global minimum value is attained if there exists a point such that is true and . Such a point is necessarily a global minimum.

If is a continuous function and the set of points satisfying the constraints is compact (closed and bounded) and nonempty, then a global minimum exists. Otherwise a global minimum may or may not exist.

Here the minimum value is not attained. The set of points satisfying the constraints is not closed:

Here the set of points satisfying the constraints is closed but unbounded. Again, the minimum value is not attained:

The minimum value may be attained even if the set of points satisfying the constraints is neither closed nor bounded:

### Local Optimization

A point is said to be a local minimum of subject to constraints if satisfies the constraints and, for some , if satisfies , then .

A local minimum may not be a global minimum. A global minimum is always a local minimum.

Here FindMinimum finds a local minimum that is not a global minimum:

### Solving Optimization Problems

The methods used to solve local and global optimization problems depend on specific problem types. Optimization problems can be categorized according to several criteria. Depending on the type of functions involved there are linear and nonlinear (polynomial, algebraic, transcendental, ...) optimization problems. If the constraints involve , you have integer and mixed integer-real optimization problems. Additionally, optimization algorithms can be divided into numeric and symbolic (exact) algorithms.

Wolfram Language functions for constrained optimization include Minimize, Maximize, NMinimize, and NMaximize for global constrained optimization, FindMinimum for local constrained optimization, and LinearProgramming for efficient and direct access to linear programming methods. The following table briefly summarizes each of the functions.

FindMinimum,FindMaximum | numeric local optimization | linear programming methods, nonlinear interior point algorithms, utilize second derivatives |

NMinimize,NMaximize | numeric global optimization | linear programming methods, Nelder-Mead, differential evolution, simulated annealing, random search |

Minimize,Maximize | exact global optimization | linear programming methods, cylindrical algebraic decomposition, Lagrange multipliers and other analytic methods, integer linear programming |

LinearProgramming | linear optimization | linear programming methods (simplex, revised simplex, interior point) |

Summary of constrained optimization functions.

Here is a decision tree to help in deciding which optimization function to use.

## Lagrange Multipliers

### How to use the Lagrange Multiplier calculator?

In mathematics, optimization techniques are available to find out the maxima and minima of the local subject and function. One of the optimization methods is the Lagrange multiplier strategy, which is used to finding the local minima and maxima of the function, which has relative constraints. It means it is subject to the constraints that have one or more equations that need to be justified by variable values. Moreover, this technique also helps you to find out the maximum and minimum of a multivariable function. For example,

### What is Lagrange Multiplier?

This optimization technique is only applicable for constraints which should be like -

g (x, y, …) = c

Where, g = multivariable function

c = constant.

When you aim to minimize or maximize a multivariable function which relates to the constraints, then follow the given steps:

- Identify a new variable λ and define a new multivariable function. Where multivariable function is known as a Lagrangian, and a new variable is represented as a Lagrange multiplier.
- Then, set the gradient of lagrangian equal to the zero vector.
- Consider each solution and add it to the multivariable function. The value for the Lagrange multiplier equals the rate of change in the maximum value of the objective function or method as the specified constraint is relaxed. The tool used for this optimization problem is known as a Lagrange multiplier calculator that solves the class of problems without any requirement of conditions to solve the data and use them to eliminate extra variables from the multivariable function.
- First, of select, you want to get minimum value or maximum value using the Lagrange multipliers calculator from the given input field.
- Then, write down the function of multivariable, which is known as lagrangian in the respective input field.
- Enter the constraint value to find out the minimum or maximum value.
- Click on the submit button, and you will get the minima or maxima value for the multivariable function.

### Use of Lagrange Multiplier Calculator

### Different constraints for multiplier

In this way, you can solve the variable function with constraints using this multiplier calculator. Added to that, you can also use this Lagrange multipliers calculator to solve the problem of three variables with one constraint. Apart from that, you can solve the optimization problem with one or two constraints using the lagrangian method calculator. As we need an efficient and speedy result, it is better to go with this kind of online tools which solves optimization problem. If there is only one constraint with the choice of two variables, consider the multiplier solver as- Maximize f (x, y)

Which is subject to: g (x, y) = 0.

The method of Lagrange multipliers calculator relies on the intuition that at a maximum value, which can nor be increasing with any neighboring point that also has another function g= 0. This method can be implemented with multiple constraints also. To get the minimum and maximum value of any variable function using a multiplier solver is that value should define in a sequence of principal minors for the second derivative for lagrangian expression.

However, this tool allows graphical representation also, which makes the process easy. Lagrange multiplier calculator changes the objective function f until its tangents the constraint function g, and the tangent points are taken as optimal points. This technique also helps to solve a production maximization problem, which gives efficient results with given conditions. Algorithm

Following are the steps that are used by the algorithm of the Lagrange multiplier calculator:

- For a multivariable function f(x,y) and a constraint which is g(x,y) = c, identify the function to be L(x, y) = f (x, y) − λ(g(x, y) − c), where λ is multiplied through the constraint.
- Now, find out the partial derivatives of the function Lx and Ly.
- In the next step, set the value of partial derivatives Lx and Ly to zero.
- Note down the details of any immediate solutions for x and y that justify Lx = 0 and Ly =0. if there is none, then move ahead to step 5.
- In this step, isolate the λ in each input equation.
- Now, solve the two λ equations and remove λ together from those equations.
- Try to reduce the equation as far as possible from the previous step. Then, you will get the relationship between x and y they shall be substituted into the constraints of this multiplier calculator.
- Substitute the equation into the constraint and algebraically solve for the remaining variables. From this point, critical points begin to be determined.
- Start to resolve for the other variable by re-assessing answers in the previous step back into the given constraint.

Make a note that if only one critical point comes, as a result, you can derive minimum or maximum value by itself. To resolve this issue while using this multiplier solver, you should study the shape of the surface and create interfaces based on the given shape and the relative location of the shape of the constraints, which is used in the Lagrange multipliers calculator. Apart from this situation, it two or more that those critical points occur, then it is an easy task to define minima and maxima values compared to the z-values.

### Lagrange Multiplier applications

You can use the Lagrange multipliers calculator for various purposes, such as to find out the maximum margin classifiers, model comparison, linear discriminant analysis, regularized least squares, and machine learning. It is also used to solve the non-linear programming problems with more complex constraint equations and inequality constraints. Still, this technique should be modified to compensate for irregular and inequality constraints and is useful to solve only small-scale problems.

Added to that, to resolve the optimization problem, the lagrangian method calculator modifies the objective function through the summation of terms of constraints.

Bottom LineHowever, its primary purpose is to find out the maximum and minimum values. This Lagrange multipliers calculator also offers the functionality of multiple constraints with multivariable functions.

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## Calculator constraints optimization with

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Harvard AM205 video 4.7 - Equality-constrained optimizationOh, damn, how huge she is. "I want to fuck your sissy :, Nina fell over on her back and, squeezing her breasts with her hands, whispered :, Here, they are waiting for you. And your baby:". Fucking Nina's luxurious breasts and at the same time squeezing Helen's tiny tits is, I can tell you, a special pleasure. - So, you want to please me, don't.

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