Thread: Max and Min

z=f(x,y)=(y-x^2)(y-2x^2)
a) Show that the point (0,0) is a critical point of f ' and that the second derivative test for a function of two variables fails at this critical point
b) Show that f has neither a maximum nor a minimum at (0,0) by proving that there are values of the parameter a for which f has a maximum at (0,0) along the parabolas y=ax^2, while for other values of parameter a, the function f has a minimum at (0,0) along the parabolas y=ax^2.
c) Show that f has a minimum at (0,0) along every straight line through(0,0)

This is really urgent to me, i need this for my test
Can anyone show me how to do? Thanks very much!

Well I hope it's not too late for your test but here goes

a) First thing I'd do is develop f so that the derivatives become easier, thus



therefore





So from this we see that f'(x,y)=0, so f is a critical point. However |f''(0,0)|=0, so the second derivative test fails.

b) let y=ax^2, thus f(x,y) becomes:





So, if a>2 or a<1, f obviously has a min. at x=0 (y is also 0 when x is 0). However if 1<a<2 then f has a max at x=0. Thus, f(x,y) has neither a max or min at (0,0)

b) let y=mx, where m is any real number. This is the family of straight lines through (0,0). f then becomes:







So, f'(0)=0, so it's a critical point, as we'd expect. Also, we have that



This is positive, and thus f has a min, for all values of m except for zero. If m=0, then,



In this case f obviously has a min at x=0. The only other case to examine is the straight line x=0, in that case, f is simply f(y)=y^2, which also has an obvious min at (0,0). Thus, f has a min at (0,0) along any straight line passing through (0,0).

Hope that helps.

thanks so much, you save my life