Lagrange multipliers

Section 14.4 Lagrange multipliers

Objectives

Recognize when substitution fails and a constraint must be handled through gradients instead.
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Use the Lagrange multiplier equation \(\Grad f=\lambda\Grad g\) to locate candidate points for constrained extrema.
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Interpret constrained optimization geometrically through level curves and tangent directions.
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Introduction goes here.

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In single-variable calculus, to find absolute extrema, we had to look at critical points and boundary points. In multivariable calculus, that boundary is now an entire shape, so we can’t just check a few points.
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To check the boundary of a shape, we can often substitute the boundary curve equation into the thing we want to optimize in order to reduce the number of variables, and then use single-variable calculus.
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We also may be able to parameterize the boundary curve to get things in terms of a single variable. This uses the Chain Rule.
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At that point, we just check all the candidates and take the highest and lowest one.
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So the big theme here is, reduce a higher dimensional problem to a lower dimensional problem.
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Need to justify this using an example where substitution goes wrong...
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Smith and Minton has an example: Find the points on the hyperbolic cylinder \(x^{2}-z^{2}-1=0\) that are closest to the origin.
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Let \(C\) be a constraint curve \(g(x,y)=k\) that lies along the surface \(z=f(x,y)\text{,}\) and suppose \(f\) has a local extreme value at \(P_{0}=(x_{0},y_{0})\) along \(C\text{.}\)
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Parametrize the curve as \(\bv{r}(t)=(x(t),y(t))\) so that \(\bv r(0)=P_{0}\text{.}\)
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Now consider the function \(h(t)=f(\bv{r}(t))\text{.}\) Then \(h\) has an extreme value at \(t=0\text{,}\) so \(h'(0)=0\text{.}\) By the Chain Rule, we have:
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\begin{align*} h'(0) \amp = \pdv{f}{x}\cdot\dv{x}{t}+\pdv{f}{y}\cdot\dv{y}{t}\\ \amp = \Grad f(P_{0})\cdot\bv{r}'(0) \end{align*}

Hence \(\Grad f\) is perpendicular to the tangent vector \(\bv{r}'(0)\text{.}\) Since \(C\) is a level curve of \(g\text{,}\) we know that \(\Grad g\) is also perpendicular to \(\bv{r}'(0)\text{,}\) and hence \(\Grad f\) and \(\Grad g\) are parallel. Hence, if \(\Grad g\neq 0\text{,}\) there exists some real number \(\lambda\) such that

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

This value \(\lambda\) is called the Lagrange multiplier.

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Perhaps would be good to do the “milkmaid” problem... its solution visually goes with the reflection property of an ellipse. And it can be explained without needing a specific curve, but then made precise using a specific curve.
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