MHB Critical Points & Extrema of Multivariable Function

harpazo
Messages
208
Reaction score
16
Find the critical points and test for relative extrema. List the critical points for which the Second Partials Test fails.

f (x, y) = x^(2/3) + y^(2/3)

Solution:

f_x = 2/[3 (x)^1/3]

f_y = 2/[3 (y)^1/3]

f_xx = -2/[9 x^(4/3)]

f_yy = -2/[9 y^(4/3)]

f_xy = 0

I set f_x and f_y to 0 and found the critical point to be
(0, 0).

To find (0, 0, 0), I evaluated f (x, y) at the point (0, 0).

Can you please tell me what (0, 0, 0) represents here? I am confused about the critical point (0, 0) and the point in space (0, 0, 0). Are they the same point?

Is this ok so far?

d = -2/[9 x^(4/3)]*-2/[9 x^(4/3)] - [0]^2

I then evaluated d at the point (0, 0) and the result is 0.
This means the test fails.

The textbook goes on to say that there is absolute minimum in this case.

Is any of this correct? Why do we have absolute minimum here and not relative minimum?
 
Physics news on Phys.org
The function:

$$f(x,y)=x^{\frac{2}{3}}+y^{\frac{2}{3}}$$

has no critical or stationary points. It has a "cusp" at $(x,y)=(0,0)$, which serves as the absolute or global minimum, and is otherwise unbounded (no global maximum).
 

Similar threads

Replies
1
Views
2K
Replies
4
Views
2K
Replies
7
Views
1K
Replies
8
Views
3K
Replies
5
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
9
Views
3K
Back
Top