1. ## Partial Derivatives

Clearly, this function has a minimum at (0,0) yet, it is not differentiable at (0,0). How then should I show that it has a minimum at (0,0)?

$\displaystyle z= \left( {x}^{2}+{y}^{2} \right) ^{ 0.5}$

First partial:

$\displaystyle \left( {x}^{2}+{y}^{2} \right) ^{- 0.5}x$

2. Originally Posted by econstudent
Clearly, this function has a minimum at (0,0) yet, it is not differentiable at (0,0). How then should I show that it has a minimum at (0,0)?

$\displaystyle z= \left( {x}^{2}+{y}^{2} \right) ^{ 0.5}$

First partial:

$\displaystyle \left( {x}^{2}+{y}^{2} \right) ^{- 0.5}x$
Well, here's a simpler question to think about .... how would you show that y = |x| has a minimum at x = 0?

3. Originally Posted by mr fantastic
Well, here's a simpler question to think about .... how would you show that y = |x| has a minimum at x = 0?
y>0 for all non-zero x. Therefore, x = 0 is the minimum. You can have a piece-wise function that shows that y is always positive for x<0 and x>0.

I guess you can use a limit?

4. ## maxima minima for 2 variable functions

Originally Posted by econstudent
y>0 for all non-zero x. Therefore, x = 0 is the minimum. You can have a piece-wise function that shows that y is always positive for x<0 and x>0.

I guess you can use a limit?
1)find partial derivative of z w.r.t. x.
2)equalise that to 0 and solve.
3)similar process for z w.r.t y.
4)you will get a point(s) (x,y) which will give either max or min values.
5) fxx.fyy-(fxy)^2>0 for max or min points
6)fxx>0 implies min value otherwise max value.

fxx=partial derivative for 2 times each time wrt x
fyy=partial derivative for 2 times wrt y
fxy=partial derivative once wrt x then wrt y.