Editing Lagrange multiplier (section)

=== Example 1 ===
[[Image:Lagrange very simple.svg|thumb|right|300px|Illustration of the constrained optimization problem&nbsp;'''1''']]

Suppose we wish to maximize <math>\ f(x,y) = x+y\ </math> subject to the constraint <math>\ x^2 + y^2 = 1 ~.</math> The [[Candidate solution|feasible set]] is the unit circle, and the [[level set]]s of {{mvar|f}} are diagonal lines (with slope −1), so we can see graphically that the maximum occurs at <math>\ \left(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)\ ,</math> and that the minimum occurs at <math>\ \left(-\tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}}\right) ~.</math>

For the method of Lagrange multipliers, the constraint is
<math display="block"> g(x,y) = x^2 + y^2-1 = 0\ ,</math>
hence the Lagrangian function,
<math display="block">\begin{align} 
\mathcal{L}(x, y, \lambda) &= f(x,y) + \lambda \cdot g(x,y) \\[4pt]
&= x + y + \lambda (x^2 + y^2 - 1)\ ,
\end{align}</math>
is a function that is equivalent to <math>\ f(x,y)\ </math> when <math>\ g(x,y)\ </math> is set to {{math|0}}.

Now we can calculate the gradient:
<math display="block">\begin{align}
\nabla_{x,y,\lambda} \mathcal{L}(x , y, \lambda) &= \left( \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial \lambda} \right ) \\[4pt]
&= \left ( 1  + 2 \lambda x, 1 + 2 \lambda y, x^2 + y^2 -1 \right)
\ \color{gray}{,}
\end{align}</math>
and therefore:
<math display="block">\nabla_{x,y,\lambda} \mathcal{L}(x , y, \lambda)=0 \quad \Leftrightarrow \quad \begin{cases} 1 + 2 \lambda x = 0 \\ 1 + 2 \lambda y = 0 \\ x^2 + y^2 -1  = 0 \end{cases}</math>

Notice that the last equation is the original constraint.

The first two equations yield
<math display="block"> x = y = - \frac{1}{2\lambda}, \qquad \lambda \neq 0 ~.</math>
By substituting into the last equation we have:
<math display="block"> \frac{1}{4\lambda^2} + \frac{1}{4\lambda^2} - 1 = 0\ ,</math>
so
<math display="block"> \lambda = \pm \frac{1}{\sqrt{2\ }}\ ,</math>
which implies that the stationary points of <math>\mathcal{L}</math> are
<math display="block">\left(\tfrac{\sqrt{2\ }}{2}, \tfrac{\sqrt{2\ }}{2}, -\tfrac{1}{\sqrt{2\ }}\right), \qquad \left(-\tfrac{\sqrt{2\ }}{2}, -\tfrac{\sqrt{2\ }}{2}, \tfrac{1}{\sqrt{2\ }} \right) ~.</math>

Evaluating the objective function {{mvar|f}} at these points yields
<math display="block">f\left(\tfrac{\sqrt{2\ }}{2}, \tfrac{\sqrt{2\ }}{2}\right) = \sqrt{2\ }\ , \qquad  f\left(-\tfrac{\sqrt{2\ }}{2}, -\tfrac{\sqrt{2\ }}{2} \right) = -\sqrt{2\ } ~.</math>

Thus the constrained maximum is <math>\ \sqrt{2\ }\ </math> and the constrained minimum is <math>-\sqrt{2}</math>.