\chapter{Methods for Constrained Nonlinear Problems}\label{C:Nonlinear_Problems}
\section{The Lagrangian Function and Lagrange Multipliers}\label{S:Lagrangian}	
\setcounter{progress_ck_counter}{0}

\noi Having focused exclusively on unconstrained NLPs in Chapters
\ref{C:UnconstrainedAlgebraicMethods} and \ref{C:Numeric_Methods}, we now turn our
attention to the constrained NLP, whose general form is given by

\begin{align}\label{E:LagrangianGeneralNLP}
\mbox{ minimize } & f(\vecx )  \\
	\mbox{ subject to} \notag\\
	g_1(\vecx) &\leq 0 \notag\\
	g_2(\vecx) &\leq 0  \notag\\
	\vdots & \notag \\
    g_m(\vecx) &\leq 0  \notag\\
   h_1(\vecx) &= 0 \notag\\
	h_2(\vecx) &= 0  \notag\\
	 \vdots & \notag \\
     h_p(\vecx) &= 0. \notag
	\end{align}
We assume that $S \subseteq \R^n$ is convex and that all functions
are differentiable mappings from $S$ to $\R$. A constraint
involving a function $g_i$, where $1 \leq i \leq m$, is of \emph{inequality-type}, whereas
 that involving $h_j$, where $1 \leq j \leq p$, is of \emph{equality-type}.
In the spirit of linear programming terminology, we say that $\vecx \in S$ is
a \emph{feasible solution} of (\ref{E:LagrangianGeneralNLP}) if it satisfies all the constraints.

\newpar
Our prototype problem is the constrained \emph{ConPro Manufacturing Company} NLP introduced in
Section \ref{SS:ConProNLP}. If $x_1$ and $x_2$ denote the number of units of material and
labor, respectively, then the NLP, stated as a minimization problem, is given by

\begin{align}\label{E:LagrangianConProConstrainedNLP}
 \mbox{minimize } f(x_1,x_2 )&= -1400 x_1^{\frac{1}{2}}x_2^{\frac{1}{3}} +350x_1+200x_2 \\
	\mbox{ subject to} & \notag\\
	g_1(\vecx) &= 350x_1+200x_2-40,000 \leq 0 \notag\\
	g_2(\vecx) &= x_1-3x_2 \leq 0  \notag\\
    g_3(\vecx) &= -x_1 \leq 0 \notag \\
    g_4(\vecx) &= -x_2 \leq 0. \notag
	\end{align}	

\newpar
We used the objective function to solve the unconstrained \CM NLP in Chapters \ref{C:UnconstrainedAlgebraicMethods} and \ref{C:Numeric_Methods}. Constraints reflect
the limitation on funds available for purchasing material and labor, the requirement of
adequate labor to produce pipes from acquired material, and sign restrictions on $x_1$
and $x_2$.

\subsection{Some Convenient Notation}

\noi Let $\vecg:S \to \R^m$ and $\vech:S \to \R^p$
be vector-valued functions defined by
$\ds \vecg(\vecx)= \begin{bmatrix} g_1(\vecx) \\ g_2(\vecx) \\ \vdots \\ g_m(\vecx) \end{bmatrix}$
and $\ds \vech(\vecx)= \begin{bmatrix} h_1(\vecx) \\ h_2(\vecx) \\ \vdots \\ h_p(\vecx) \end{bmatrix}$.
Then
(\ref{E:LagrangianGeneralNLP}) can be written more compactly as

\begin{align}\label{E:LagrangianGeneralNLPVectorForm}
 \mbox{ minimize }  & f(\vecx )  \\
	\mbox{ subject to} & \notag\\
	\vecg(\vecx) &\leq \zerovec \notag\\
\vech(\vecx) &= \zerovec. \notag
	\end{align}	

\newpar
The function $\vecg$ is a vector-valued function whose components are of the type
discussed in Section \ref{S:diff_Gradient}. Consequently, we define
$\vecg$ to be differentiable at  $\vecx_0$ in $S$ if and only if each component function
$g_i$, where $1 \leq i \leq m$, is differentiable there. When this occurs, we write

\begin{equation}\label{E:KKTdiff_def}
\vecg(\vecx )= \vecg(\vecx_0 )+ \jacobian_{\vecg}(\vecx_0)( \vecx-\vecx_0 )+ \|\vecx -\vecx_0\|\vecR(\vecx;\vecx_0),
\end{equation}
where all components of $\vecR(\vecx;\vecx_0) \in \R^m$ tend to $\zerovec$ as $\vecx \to \vecx_0$ and
where $\jacobian_{\vecg}(\vecx_0)$ is an $m$-by-$n$ matrix, called
the \emph{Jacobian}\index{Jacobian} of $\vecg$ evaluated at $\vecx_0$. It is the matrix whose $i$th row, $1 \leq i \leq m$, is given by the gradient vector transpose,
$\nabla g_i(\vecx_0)^t$.

\newpar
The Jacobian matrix yields the linear approximation
\begin{equation}\label{E:LagrangianLocalApproximation}
\vecg(\vecx )\approx \vecg(\vecx_0 )+ \jacobian_{\vecg}(\vecx_0) ( \vecx-\vecx_0 ) \mbox{  for  } \vecx \approx \vecx_0.
\end{equation}

\newpar
Naturally, the preceding analysis can be modified when $\vecg$ is
replaced by $\vech$, in which case $\jacobian_{\vech}(\vecx_0)$ is a $p$-by-$n$ matrix.

\newpar
In Maple, the Jacobian is computed using the \mf{Jacobian} command\index{Maple!\mf{Jacobian}}, located in the
\mf{VectorCalculus} package. Its general form is given by
\begin{equation*}
\mbox{ \mf{Jacobian(\ma{vector of expressions,variable list})}, }
\end{equation*}
and it computes the Jacobian of \ma{vector of expressions} with respect to the variables given in \ma{variable list}. The output of the command
is a matrix. Here is syntax illustrating how to compute the Jacobian of the function
$\vecg$, where $\vecg$ is formed using the four constraints of the
\emph{ConPro Manufacturing Company} NLP
(\ref{E:LagrangianConProConstrainedNLP}). In this example, we construct the Jacobian
as a function of the vector, $\ds \vecx = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.

\begin{maplelist}
\item \mf{with(VectorCalculus):}
\item \mf{g1:=(x1,x2)->350*x1+200*x2-40000:}
\item \mf{g2:=(x1,x2)->x1-3*x2:}
\item \mf{g3:=(x1,x2)->-x1:}
\item \mf{g4:=(x1,x2)->-x2:}
\item \mf{g:=(x1,x2)-><g1(x1,x2),g2(x1,x2),g3(x1,x2),g4(x1,x2)>:}
\item \mf{g(x1,x2);}
\mo{(350x_1+200x_2-40000)\vece_{x_1} + (x_1-3x_2)\vece_{x_2}  -x_1\vece_{x_3}  -x_2\vece_{x_4}}
\item \mf{Jg := unapply(Jacobian(g(x1, x2), [x1, x2]), [x1, x2]):}
\item \mf{Jg(x1,x2);}
\mo{\begin{bmatrix} 350 & 200 \\ 1 & -3 \\ -1 & 0 \\ 0 & -1 \end{bmatrix}}
\end{maplelist}

\subsection{The Karush-Kuhn-Tucker Theorem}

\noi Key to solving constrained NLPs is the \emph{Lagrangian function}\index{Lagrangian function},
a single function that incorporates the objective as well as all constraints.

\begin{definition}\label{Th:LagrangianFunctionDefinition}
The Lagrangian function corresponding to NLP (\ref{E:LagrangianGeneralNLP}) is the function
$L:S \times \R^{m+p} \to \R$ defined by
\begin{align}\label{E:LagrangianFunctionFormula}
L(\vecx, \lambdavec,\muvec)&=f(\vecx)+\sum_{i=1}^m \lambda_i g_i(\vecx)
+\sum_{j=1}^p \mu_jh_j(\vecx)
 \\
&=f(\vecx)+\lambdavec^t\vecg(\vecx)+ \muvec^t \vech(\vecx), \notag
\end{align}
where $\vecx$ belongs to $\R^n$ and where $\lambdavec$ and $\muvec$
are vectors in $\R^m$ and $\R^p$, respectively, denoted by
$\ds \lambdavec=\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_m \end{bmatrix}$
and $\ds \muvec=\begin{bmatrix} \mu_1 \\ \mu_2 \\ \vdots \\ \mu_p \end{bmatrix}$.
\end{definition}


\newpar
Note that in each product,
$\ds \lambdavec^t\vecg(\vecx)$ and
$\ds \muvec^t\vech(\vecx)$,
there exists a natural correspondence between each constraint of the NLP
(\ref{E:LagrangianGeneralNLP}) and a component of $\lambdavec$ or $\muvec$. In particular,
a constraint of inequality-type (respectively, of equality-type) corresponds
to a component of $\lambdavec$ (respectively, to a component of $\muvec$).

\newpar
Our subsequent investigation focuses on the Lagrangian function. Using this
tool, we obtain important information regarding solutions of
(\ref{E:LagrangianGeneralNLP}). The overall process closely resembles
that from Sections \ref{S:diff_Gradient}-\ref{S:optimality_sufficient_SOC} in that we
first determine necessary conditions $\vecx_0$ in $\R^n$
to be a solution of (\ref{E:LagrangianGeneralNLP}). Then we establish sufficient conditions for
$\vecx_0$ to be a solution of this NLP. These tools, which were developed in the mid 1900s by
Fritz John\index{John,F.}, William Karush\index{Karush, W.}, Harold William Kuhn,\index{Kuhn, H.W.}
and Albert William Tucker\index{Tucker, W.}, form the basis of constrained optimization theory.

\newpar
Throughout this discussion, unless stated otherwise, we use the notation
$\ds
\nabla L(\vecx,\lambdavec,\muvec)$ to denote the gradient of $L$ with respect to the components of $\vecx$. A useful
formula that expresses $\ds \nabla L(\vecx,\lambdavec,\muvec)$ in terms of the functions
that define $L$ is given by
\begin{align}\label{E:LagrangianFunctionGradient}
\nabla L(\vecx,\lambdavec,\muvec)&=
\nabla f(\vecx_0) + \sum_{i=1}^m \lambda_{0,i} \nabla g_j(\vecx_0) +
\sum_{j=1}^p \mu_{0,j} \nabla h_j(\vecx_0) \\
&=\nabla f(\vecx) + \jacobian_{\vecg}(\vecx)^t \lambdavec
+ \jacobian_{\vech}(\vecx)^t \muvec.\notag
\end{align}
\noi
We now state an important necessary condition, in terms of the Lagrangian, for a feasible solution of
 (\ref{E:LagrangianGeneralNLPVectorForm}) to be optimal.

\begin{theorem}\label{Th:KKTTheorem}\index{Karush-Kuhn-Tucker Theorem}
Consider the NLP (\ref{E:LagrangianGeneralNLPVectorForm}) and assume that $f$ and all component
functions of $\vecg$ and $\vech$ are continuously differentiable on
$S$. Suppose that $\vecx_0$ is a solution of the NLP, and let $I$ denote the
set of indices corresponding to \emph{binding}\index{binding constraint} inequality constraints. In other words
\begin{equation}\label{E:IsetDefinition}
I= \left\{ i \mid 1 \leq i \leq m \mbox{ and } g_i(\vecx_0)=0 \right\}.
\end{equation}

\noi
Let $V$ be the set of gradient vectors corresponding to both equality constraints and binding inequality constraints, each of which is evaluated at $\vecx_0$. In other words,
\begin{equation*}
V= \ds \left\{\nabla g_i(\vecx_0) \mid i \in I \right\} \cup \left\{\nabla h_j(\vecx_0) \mid 1 \leq j \leq p \right\}.
\end{equation*}
If $V$ forms a linearly independent set of vectors, a provision we refer to
as the the \emph{regularity condition}\index{regularity condition}, then there exist unique vectors, $\lambdavec_0 \in \R^m$ and
$\muvec_0 \in \R^p$, having the following three properties:
\begin{align}
\lambdavec_0 &\geq \zerovec \label{E:NonnegativityFritzJohnFormula} \\
\nabla L(\vecx_0,\lambdavec_0,\muvec_0) &=\nabla f(\vecx_0) + \jacobian_{\vecg}(\vecx_0)^t \lambdavec_0
+ \jacobian_{\vech}(\vecx_0)^t \muvec_0 =\zerovec \label{E:LagrangianFritzJohnFormula} \\
\lambda_{0,i}g_i(\vecx_0) &=0 \quad \mbox{for } 1 \leq i \leq m. \label{E:LagrangianFritzJohnComplementarySlackness}
\end{align}
\end{theorem}

\newpar
Theorem \ref{Th:KKTTheorem} is often referred to as the Karush-Kuhn-Tucker Theorem (or Karush-Kuhn-Tucker Necessary Conditions Theorem). We omit its proof, which can be found in more
advanced texts \cite{Avriel},\cite{Bazaraa}. Instead, we take a closer look at its meaning.

\be
\item Entries of the vectors
$\lambdavec_0$ and $\muvec_0$ are referred to as the \emph{Lagrange multipliers}\index{Lagrange multiplier}
associated with the NLP (\ref{E:LagrangianGeneralNLPVectorForm}), and the vectors
$\lambdavec_0$ and $\muvec_0$ themselves as the \emph{multiplier vectors}\index{Lagrange multiplier!vector}.

\item Any feasible value $\vecx_0$ for which conditions (\ref{E:NonnegativityFritzJohnFormula})-(\ref{E:LagrangianFritzJohnComplementarySlackness})
hold is called a \emph{Karush-Kuhn-Tucker}, or \emph{KKT-point}\index{Karush-Kuhn-Tucker (KKT) point}
of NLP (\ref{E:LagrangianGeneralNLPVectorForm}). If the regularity condition also holds at $\vecx_0$, we say the KKT point is regular.

\item The regularity condition cannot be overlooked. If we omit this requirement altogether, the multiplier vectors satisfying conditions (\ref{E:NonnegativityFritzJohnFormula})-(\ref{E:LagrangianFritzJohnComplementarySlackness}) need
    not exist in the first place, or, if they do exist, they need not be unique. Exercises 2 and 3 establish these facts.

\item An inequality constraint of the form
$\ds g_i(\vecx)\leq 0$ corresponds to the sign restriction $\lambda_{0,i} \geq 0$, whereas
$\ds h_j(\vecx)= 0$ corresponds to a multiplier
$\mu_{0,j}$ that is \emph{unrestricted in sign}\index{unrestricted in sign}. Results from Section
\ref{S:Duality}, such as those summarized in Table \ref{Ta:Duality_general_dual_formulation},
suggest
an intriguing connection between the multiplier vectors, $\lambdavec_0$ and $\muvec_0$, and
duality.
\item This connection with duality is reinforced if we observe that condition (\ref{E:LagrangianFritzJohnComplementarySlackness}) corresponds to the complementary
 slackness property, Theorem \ref{Th:Duality_complementary_slackness}, from Section \ref{S:Duality}. Stated in simple terms, at an optimal solution of the NLP, $\vecx_0$, which satisfies the
 theorem's hypothesis, each multiplier,
 $\lambda_{0,i}$, is zero, or the corresponding constraint, $\vecg_i(\vecx)$,
 is \emph{binding}\index{binding constraint}. As a result of this fact, in subsequent discussions we shall refer to (\ref{E:LagrangianFritzJohnComplementarySlackness})
as the \emph{complementary slackness conditions}\index{complementary slackness}. Since $\lambda_{0,i} \geq 0$ and $g_i(\vecx_0) \leq 0$ for
\newline $1 \leq i \leq m$, complementary slackness can be expressed in matrix-vector product
form as $\ds \lambdavec_0^t \vecg(\vecx_0)=0$.

\item As a consequence of conditions (\ref{E:LagrangianFunctionGradient}),
    (\ref{E:LagrangianFritzJohnFormula}), and the fact $\lambda_{0,i}=0$ for $i$
    not belonging to $I$, we have,
\begin{align}\label{E:LagrangianLinearIndependence}
\zerovec&= \nabla L(\vecx_0,\lambdavec_0,\muvec_0) \\
&= \nabla f(\vecx_0) + \jacobian_{\vecg}(\vecx_0)^t \lambdavec_0
+ \jacobian_{\vech}(\vecx_0)^t \muvec_0 \notag \\
&= \nabla f(\vecx_0) + \sum_{i=1}^m \lambda_{0,i} \nabla g_j(\vecx_0) +
\sum_{j=1}^p \mu_{0,j} \nabla h_j(\vecx_0) \notag \\
&= \nabla f(\vecx_0) + \sum_{i \in I} \lambda_{0,i} \nabla g_i(\vecx_0) +
\sum_{j=1}^p \mu_{0,j} \nabla h_j(\vecx_0).
 \notag
\end{align}

\noi
In other words, $\ds \nabla f(\vecx_0)$ is a linear combination of the vectors in $V$,
with weights determined by the Lagrange Multipliers.
\ee

%************************************************************
%***********************************************************

\newpar
To investigate these ideas in the context of a particular example and to understand how the solution of an NLP determines the Lagrange multiplier vectors, we consider
the \emph{ConPro Manufacturing Company} NLP (\ref{E:LagrangianConProConstrainedNLP}). In the next section, we establish that the optimal solution is given
by
\begin{equation}\label{E:LagrangianConProSolution}
\vecx_0 = \begin{bmatrix} 480/7 \\ 80 \end{bmatrix} \approx \begin{bmatrix}  68.57 \\ 80 \end{bmatrix},
\end{equation}
with corresponding objective value, $\ds f(\vecx_0) = -9,953.15$. Thus,
using approximately 68.57 units of material and 80 units of labor, \emph{ConPro Manufacturing Company} has
a maximum profit of \$9,953.15, given the constraints dictated in (\ref{E:LagrangianConProConstrainedNLP}).
Observe that this solution differs from that of the unconstrained version of the problem,
which we solved in Section \ref{S:optimality_sufficient_SOC} and which
had solution $\ds \vecx_0=\begin{bmatrix} 784/9 \\ 2744/27 \end{bmatrix}
\approx \begin{bmatrix} 87.11 \\ 101.63 \end{bmatrix}$
with corresponding profit approximately equal to \$10,162.96.

\newpar
A simple calculation shows that at the optimal solution, only the first
constraint, $g_1(\vecx)=350x_1+200x_2-40,000 \leq 0$ is binding. Thus,
$\ds \nabla g_1(x_0)=\begin{bmatrix} 350 \\ 200 \end{bmatrix}$ trivially satisfies
the regularity condition, so by
(\ref{E:LagrangianFritzJohnComplementarySlackness}),
$\ds \lambda_{0,2}=\lambda_{0,3}=\lambda_{0,4}=0$.

\newpar The Lagrangian has its gradient given by
\begin{align}\label{E:LagrangianCMCLagrangianGradient}
\nabla L(\vecx,\lambdavec)&=\nabla f(\vecx) + \jacobian_{\vecg}(\vecx)^t \lambdavec \\
&= - \begin{bmatrix} \frac{ 700x_2^{\frac{1}{3} }}{\sqrt{x_1}}+350
\\  \frac{1400\sqrt{x_1}  }{ 3x_2^{\frac{2}{3} }} +200  \end{bmatrix}
+ \begin{bmatrix} 350 & 1 & -1 & 0 \\ 200 & -3 & 0 & -1 \end{bmatrix}
\begin{bmatrix} \lambda_{1} \\ \lambda_{2} \\ \lambda_{3} \\ \lambda_{4} \end{bmatrix} \notag \\
&= \begin{bmatrix} -\frac{ 700x_2^{\frac{1}{3} }}{\sqrt{x_1}}+350 + 350 \lambda_1 + \lambda_2 - \lambda_3 \\  -\frac{1400\sqrt{x_1}  }{ 3x_2^{\frac{2}{3} }} + 200 + 200\lambda_1 - 3 \lambda_2
-\lambda_4 \end{bmatrix}. \notag
\end{align}

\noi
Evaluating $\nabla L$ at $\ds \vecx_0=\begin{bmatrix} 480/7 \\ 80 \end{bmatrix}$
and $\ds \lambda_{0,2}=\lambda_{0,3}=\lambda_{0,4}=0$ and setting each component of the result
to zero, we obtain
\begin{equation}\label{E:LagrangianCMCLambda1Value}
\lambda_{0,1} = \sqrt{\frac{7}{120}}\sqrt[3]{80}-1 \approx .04.
\end{equation}

\newpar
These results demonstrate that at the optimal solution of NLP (\ref{E:LagrangianConProConstrainedNLP}), \newline
$\ds \vecx_0=\begin{bmatrix} 480/7 \\ 80 \end{bmatrix}$, with the Lagrange
multiplier vector given by $\lambdavec_0 \approx \begin{bmatrix} .04 \\ 0 \\ 0 \\ 0 \end{bmatrix}$.
In light of the linear dependence relation spelled out by
(\ref{E:LagrangianLinearIndependence}), we see that
\begin{align*}
\zerovec &= \nabla f(\vecx_0) + \jacobian_{\vecg}(\vecx_0)^t \lambdavec_0 \notag \\
&= \nabla f(\vecx_0) + \lambda_{0,1} \nabla g_1(\vecx_0) \notag \\
&\approx \nabla f(\vecx_0) + .04 \nabla g_1(\vecx_0) .\notag
\end{align*}

\noi
Thus, for the \emph{ConPro Manufacturing Company} NLP, at the optimal solution, $\vecx_0$ the gradients of $f$ and the first constraint function, $g_1$, are scalar multiples of one another, where the scaling
factor is approximately $-.04$.

\newpar
Figure \ref{Fi:f09x01x01} illustrates the feasible region for the \emph{ConPro Manufacturing Company} NLP,
the optimal solution, $\ds \vecx_0 \approx \begin{bmatrix} 68.57 \\ 80  \end{bmatrix}$,
$\nabla f(\vecx_0)$, and $\nabla g_1(\vecx_0)$. Vectors have been scaled for
purposes of visualization.

\begin{figure}[h]
\centering
\psfrag{x1_axis_label}[Bc][Bc]{$x_1$}
\psfrag{x2_axis_label}[Bc][Bc]{$x_2$}
\psfrag{x0}[Bc][Bc]{$\vecx_0$}
\psfrag{del_f}[Bc][Bc]{$\nabla f$}
\psfrag{del_g}[Bc][Bc]{$\nabla g_1$}
\includegraphics[width=.7\linewidth]{images/f09x01x01.eps}
\caption{Feasible region of the \emph{ConPro Manufacturing Company} NLP illustrating objective and constraint gradients at
the solution, $\vecx_0$.}
\label{Fi:f09x01x01}
\end{figure}

\subsection{Interpreting the Multiplier}\index{Lagrange multiplier!interpretation}

\noi
The multiplier, $\lambda_{0,1} \approx .04$ in the \emph{ConPro Manufacturing Company} NLP has a simple economic interpretation. First, observe that the units
of the Lagrangian function in (\ref{E:LagrangianCMCLagrangianGradient}) are the same as the units of $f$, namely dollars (of profit). Since the units of $g_1$, the only binding constraint in the solution,
are dollars (of available funds for material and labor), the units of  $\lambda_{0,1}$ are dollar of profit per dollar of available material and labor. Put another way, the multiplier $\lambda_{0,1}$
is the instantaneous rate of change of the objective, $f$, with respect to the constraint function,
$g_1$. This means that if $g_1$ changes by a small amount, small enough so that it remains a binding constraint in the new solution, we should notice the value of $f$ at the optimal solution change by an amount approximately equal to the change in $g_1$, multiplied by $\lambda_{0,1}$.

\newpar For example, suppose the funds available to \CM for material and labor increases
 by \$100 from \$40,000 to \$40,100. Then the new first constraint becomes
 $\ds 350x_1+200x_2 \leq \mbox{40,100}$, or, equivalently, $g_1$ in (\ref{E:LagrangianConProConstrainedNLP})
changes to
\begin{equation}\label{E:LagrangianNew_g1}
g_1(\vecx) = 350x_1+200x_2-\mbox{40,100}.
\end{equation}
In terms of Figure \ref{Fi:f09x01x01}, such a change corresponds to moving slightly up and to the right the segment on the boundary of the feasible region
that passes through $\ds \vecx_0=\begin{bmatrix} 480/7 \\ 80 \end{bmatrix}$. The multiplier,
$\lambda_{0,1} \approx .04$, indicates, that, under such a change, we should expect to see the objective
value in the solution
of the new NLP to change by $\Delta f \approx .04 (-100)=-4$. This is a fairly good approximation,
for if we replace the original first constraint by that in (\ref{E:LagrangianNew_g1}) and solve
the resulting NLP (with Maple), we obtain a new optimal objective value of
$-9,957.21$. This value, less that of  $\ds f(\vecx_0) = -9,953.15$, is approximately -4.06. Thus, with \$100 more funds available for capital and
labor, \CM's profit increases by a mere \$4.06.

\newpar
In essence, we can view the multiplier as playing a role in sensitivity analysis when we increase the
bound on the right-hand side of a constraint. In fact, if we recall the concept
of a \emph{shadow price}\index{shadow price} introduced in Section \ref{S:Sensitivity_constraint_change}, we see
that the multiplier generalizes this idea to the nonlinear setting. What is more, shadow prices
in the LP setting equalled the slack variable coefficients in the top row of the tableau, which
in turn corresponded to the decision variables in the LP's dual formulation. That Lagrange multipliers are somehow connected to dual decision variables for an LP becomes the focus of further investigation in Exercise (5).

\section*{Exercises Section \protect\ref{S:Lagrangian}}

\addcontentsline{toc}{subsection}{Exercises}

\input{Exercises/Exercises_Lagrangian}	
\endinput








