Lagrange's Method of Multipliers/Examples/Arbitrary Example 1

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Examples of Use of Lagrange's Method of Multipliers

Find the maximum $M$ of the function $u: \R^2 \to \R$ defined as:

$\forall \tuple {x, y} \in \R^2: \map u {x, y} = x y$

subject to the constraint:

$\text C: \quad x + y = 1$


Solution

The maximum $M$ is

$M = \dfrac 1 4$

at $\tuple {x, y} = \tuple {\frac 1 2, \frac 1 2}$.


Proof

We write:

$L = x y + \lambda \paren {x + y - 1}$

Differentiation with respect to $x$, $y$ and $\lambda$ and equating to zero gives:

\(\text {(1)}: \quad\) \(\ds y + \lambda\) \(=\) \(\ds 0\)
\(\text {(2)}: \quad\) \(\ds x + \lambda\) \(=\) \(\ds 0\)
\(\text {(3)}: \quad\) \(\ds x + y - 1\) \(=\) \(\ds 0\)

This leads to:

\(\ds x\) \(=\) \(\ds \dfrac 1 2\) $(2) - (1) + 3$ gives $2 x - 1 = 0$
\(\ds y\) \(=\) \(\ds \dfrac 1 2\) by symmetry and the above
\(\ds \lambda\) \(=\) \(\ds -\dfrac 1 2\) substituting for $x$ or $y$ in $(1)$ or $(2)$

giving:

$u = \dfrac 1 4$

This is the only stationary point for $L$.

$\Box$


We need to confirm it is indeed a maximum for this constraint:

$\text C: \quad x + y = 1$

We check at $\tuple {1, 0}$ and $\tuple {0, 1}$, both of which obey $\text C$, and confirm that $u = 0$ at both points.

So it goes:

up from $0$ at $\tuple {1, 0}$ to $\dfrac 1 4$ at $\tuple {\frac 1 2, \frac 1 2}$
back down again from $\dfrac 1 4$ at $\tuple {\frac 1 2, \frac 1 2}$ to $0$ at $\tuple {0, 1}$.


There is only one stationary point for $L$.

Hence at $\dfrac 1 4$ it must be a maximum.

$\blacksquare$


Sources

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