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Lagrange Multipliers

Find the extreme values of the function $f(x,y) = 2x^2 +y^2$ on the circle $x^2 + y^2 = 1$.


\begin{maximasession} f(x,y) := 2 * x^2 + y^2; g: x^2 + y^2; \maximaoutput* \i1.... ...in{:=}2 x^2+y^2 \\ \i2. g: x^2 + y^2; \\ \o2. y^2+x^2 \\ \end{maximasession}

We set up the system of equations grad(f) = h * grad(g), g = 1. (We do not use "lambda" because that name is already reserved for an existing function in Maxima.)


\begin{maximasession} eq1: diff(f(x,y), x) = h * diff(g, x); eq2: diff(f(x,y), y... ...o4. 2 y=2 h y \\ \i5. eq3: g = 1; \\ \o5. y^2+x^2=1 \\ \end{maximasession}

Now we solve the system for x, y, and h.


\begin{maximasession} solve([eq1, eq2, eq3], [x, y, h]); \maximaoutput* \i6. sol... ...1 , h=1 \right] , \left[ x=0 , y=1 , h=1 \right] \right] \\ \end{maximasession}

We see that the extreme values lie among $(1,0)$, $(-1,0)$, $(0,-1)$, $(0,1)$.


\begin{maximasession}[f(1,0), f(-1,0), f(0,-1), f(0,1)]; \maximaoutput* \i7. [f(... ... f(0,-1), f(0,1)]; \\ \o7. \left[ 2 , 2 , 1 , 1 \right] \\ \end{maximasession}

So the minima occur at $(1,0)$ and $(-1,0)$; the maxima occur at $(0,-1)$ and $(0,1)$.




G. Jay Kerns 2009-12-01