What is the critical point of f(x,y) = 4x^6 + 5y^6 + 6?

Question

What is the critical point of f(x,y) = 4x^6 + 5y^6 + 6 and determine whether its a local max, min or saddle point. I got stuck on finding the max, min or saddle point and need some help.

Here is my working:

fx(x,y) = 24x^5

fy(x,y) = 30y^5

when fx(x,y) = 0,

24x^5 = 0

x = 0

when fy(x,y) = 0,

30x^5 = 0

y = 0

critical point (0, 0)

then,

fxx(x,y) = 120x^4

fyy(x,y) = 150y^4

fxy(x,y) = fyx(x,y) = 0

D = 18000x^4y^4

base on my notes,

if D > 0, either local max if fxx < 0 or local min if fxx > 0

if D < 0, saddle point

i subtitute point (0,0) to D and i get 0....

how can that be??

If I plot a graph, I would know that the point (0,0) would be the local minimum. Which step did I did wrongly?

Ben · Accepted Answer

You've done nothing incorrectly so far.  If D=0, then you need another approach to decide whether it is a max, min, or saddle.  In this case, notice that each term in f(x,y) has an even power.  Thus f(x,y)>=6 for all x and y, and f(x,y)=6 only at (0,0).  So (0,0) is the global minimum, hence certainly a local one.

(The plot of f looks a bit like a squared-off elliptical paraboloid.)

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What is the critical point of f(x,y) = 4x^6 + 5y^6 + 6?

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