Constrained Extrema Homework: Find Min & Prove Limit of 0

[teleport]

Homework Statement

These are two problems from my assignment that are due tomorrow! Plz, help as you can.

1)Show that f(x,y) = (y - x)(y - 3x^2) attains a local minimum on every straight line through the origin and that this occurs at (0,0). Does f have a local minimum at (0,0)?

2)Assuming that f_x and f_y both exist at (x,y), prove that if

$\lim_{(h,k)\rightarrow(0,0)}\dfrac{f(x+h, y) - f(x,y) - f_{x}(x,y)h - f_{y}(x,y)k}{\sqrt{h^2 + k^2}}$

exists, the limit is 0.

Homework Equations

The Attempt at a Solution

For the first one I tried Lagrange's multipliers but became a mess with all the algebra, leaving me with the impression that this might be done more easily using something that I might have missed in class.

For the 2nd one, honestly, no clue. I left my assignment for the last day, because I had so many other stuff to do. Now I'm really in trouble. Thanks for all your help.

 

[teleport]

#2 is done! It wasn't as terrible as I thought it would be. #1 still hurts. I just don't know what to do. Help please.

 

[HallsofIvy]

Homework Statement

These are two problems from my assignment that are due tomorrow! Plz, help as you can.

1)Show that f(x,y) = (y - x)(y - 3x^2) attains a local minimum on every straight line through the origin and that this occurs at (0,0). Does f have a local minimum at (0,0)?

2)Assuming that f_x and f_y both exist at (x,y), prove that if

$\lim_{(h,k)\rightarrow(0,0)}\dfrac{f(x+h, y) - f(x,y) - f_{x}(x,y)h - f_{y}(x,y)k}{\sqrt{h^2 + k^2}}$

exists, the limit is 0.

Homework Equations

The Attempt at a Solution

For the first one I tried Lagrange's multipliers but became a mess with all the algebra, leaving me with the impression that this might be done more easily using something that I might have missed in class.

I don't see why. The gradient of $(y - x)(y - 3x^2)$ is $[(-y-3x^2)-6x(y-x)]\vec{i}+ [2y-x-3x^2]\vec{j}$ while we can write the condition y= mx as G(x,y)= y- mx= 0 and its gradient is $-m\vec{i}+ \vec{j}$. Lagranges multiplier method gives you two equations: $(y-3x^2)-6x(y-x)= -\lambda m$ and $2y-x-3x^2= \lambda$. Divide one equation by the other to get rid of $\lambda$ (that's a standard technique), then replace y with mx and it reduces to a quadratic equation to solve for x (in terms of m, of course).

 

[teleport]

But how do you make sure that (0,0) gives the min?