Using the method of Lagrange Multipliers, find the maximum and minimum values of \(f(x,y) = xy\) on the ellipse \(x^2 + 4y^2 = 8\).
Solution
Let \[ g(x,y)=x^2+4y^2. \] At an extremum on \(g=8\), Lagrange multipliers give \[ (y,x)=\lambda(2x,8y), \] so \[ y=2\lambda x,\qquad x=8\lambda y. \] Neither \(x\) nor \(y\) can be zero at such a point, so substitution gives \[ 1=16\lambda^2. \] Hence \(\lambda=\pm1/4\).
If \(\lambda=1/4\), then \(y=x/2\). The constraint becomes \[ x^2+4(x/2)^2=2x^2=8, \] so \(x=\pm2\) and \(y=\pm1\) with matching signs. Thus \(xy=2\).
If \(\lambda=-1/4\), then \(y=-x/2\), giving the points \((2,-1)\) and \((-2,1)\) and hence \(xy=-2\).
Therefore the maximum value is \(2\) and the minimum value is \(-2\).