In class, we found the maximum and minimum values of f(x,y)=(x^4)+3(y^4)+2 given (x^2)+(y^2)=1.  This corresponds to finding the highest and lowest point of the blue curve in the following picture.

Show[
Plot3D[(x^4)+3(y^4)+2,{x, -1.5, 1.5},{y, -1.5,1.5},Mesh->None,PlotStyle->Opacity[0.3]],
ParametricPlot3D[{{Cos[t], Sin[t],((Cos[t])^4)+3((Sin[t])^4)+2},{Cos[t], Sin[t],0}},{t, 0, 2Pi}]
,ImageSize->Large]

Using Lagrange multipliers, we found that the test points gave us values 5, 3, and (11/4). The level curves of f at these values along with the constraint curve are graphed below.

Show[ContourPlot[(x^2)+(y^2)==1,{x, -1.5, 1.5}, {y, -1,1},ColorFunction->Function[{x,y},Red]],
ContourPlot[{(x^4)+3(y^4)+2 ==(11/4),(x^4)+3(y^4)+2 ==5, (x^4)+3(y^4)+2 ==3},{x, -2, 2},{y, -2,2}]]

We see that each of the level curves of of f has points of tangency with the 	 As the value k in the level curve f(x,y)=k decreases, the rounded circles representing these level curves also shrink.  Each of the graphed level curves have a point of tangency with the red ellipse.  

If the “radius” of the yellow level curve f(x,y)=5 were increased a little, it would no longer intersect the level curve (x^2)+(y^2)=1; this agrees with the math that showed 5 was the maximum value of f on this curve.  

If the “radius” of the blue level curve f(x,y)=(11/4) were decreased a little, it would no longer intersect the level curve (x^2)+(y^2)=1; this agrees with the math that showed (11/4) was the minimum value of f on this curve. 

If the “radius” of the green level curve f(x,y)=3 were changed, it would still intersect the level curve (x^2)+(y^2)=1; this agrees with the conclusion that 3 was neither the maximum or the minimum value.