Finding critical points of functions of two variables

[th4m4ster]

Hey,

I have trouble finding the critical points for this function:

z = f(s,t) = (1+s)(1+t)(s+t)

I get that
s' = (1+t) (t+2s+1) = 0
t' = (1+s) (s+2t+1) = 0

So t= -1 and s= -1 is a C.P.

How do i solve for the others?

Any help is greatly appreciated thanks!

 

[tiny-tim]

welcome to pf!

hey th4m4ster! welcome to pf!

s' = (1+t) (t+2s+1) = 0
t' = (1+s) (s+2t+1) = 0

So t= -1 and s= -1 is a C.P.

How do i solve for the others?

(you meant ∂z/∂s and ∂z/∂t, not s' and t' )

solve for (1+t) = 0 and (s+2t+1) = 0,

and for the other two combinations

 

[th4m4ster]

hey th4m4ster! welcome to pf!


(you meant ∂z/∂s and ∂z/∂t, not s' and t' )

solve for (1+t) = 0 and (s+2t+1) = 0,

and for the other two combinations

Hi Tiny-Tim,

thanks for the quick reply!

I do understand I have to solve for 1+t = 0 as well as s+2t+1 =0
and 1+s = 0 and t+2s+1=0.

The difficulty I am having is with s+2t+1 =0 and t+2s+1=0. The two others are clearly s=t=-1

Therefore (-1,-1) is a solution. Now, how do I solve for the two others? I thought of replacing t=-1 in

t+2s+1=0

and s=-1 in s+2t+1 =0. Although that gives me 4 points in the end, it seems kind of weird because I end up with only saddle points, and no local min. I don't have the answers for this question, but I am pretty sure there is a local minimum in this question, because one part of the question asks what is the minimum value of z.

 

[th4m4ster]

I think I got it

I ended up with (-1,-1) , (-1,1) , (1,-1) , (1,1)

and the three first are saddle points while the last one is a local minimum. Anyone correct me if the person attempts the problem. Thanks

 

[HallsofIvy]

Hey,

I have trouble finding the critical points for this function:

z = f(s,t) = (1+s)(1+t)(s+t)

I get that
s' = (1+t) (t+2s+1) = 0
t' = (1+s) (s+2t+1) = 0

I assume you mean that z_s and z_t rather than s' and t'.
From the first equation, either t= -1 or t+ 2s+ 1= 0. If t= -1, the second equation becomes (1+ s)(s- 1)= 0 so we have either s= -1 or s= 1. Two critical points are s= -1, t=-1 and s= 1, t= -1.

From the second equation, either s= -1 or s+ 2t+ 1= 0. If s= -1, the first equation becomes (1+ t)(t- 1)= 0 so either t= -1 or t= 1. Two critical points are s= -1, t= -1, which we had before, and s= -1, t= 1.

If neither s= -1 or t= -1, then we must have t+ 2s+ 1= 0 and s+ 2t+ 1= 0. From the first equation, t= -2s- 1 so the second equation becomes s+ 2(-2s- 1)+ 1= s- 4s- 2+ 1= -3s-1= 0 and s= -1/3. In that case, t= -2(-1/3)- 1= 2/3- 1= -1/3.

The critical points are s= -1, t= -1; s= 1, t= -1; s= -1, t= 1; and s= -1/3, t= -1/3.

s= t= 1 is NOT a critical point: (1+ 1)(1+ 2+ 1)= 2(4)= 8 not 0

So t= -1 and s= -1 is a C.P.

How do i solve for the others?

Any help is greatly appreciated thanks!

 

[th4m4ster]

I assume you mean that z_s and z_t rather than s' and t'.
From the first equation, either t= -1 or t+ 2s+ 1= 0. If t= -1, the second equation becomes (1+ s)(s- 1)= 0 so we have either s= -1 or s= 1. Two critical points are s= -1, t=-1 and s= 1, t= -1.

From the second equation, either s= -1 or s+ 2t+ 1= 0. If s= -1, the first equation becomes (1+ t)(t- 1)= 0 so either t= -1 or t= 1. Two critical points are s= -1, t= -1, which we had before, and s= -1, t= 1.

If neither s= -1 or t= -1, then we must have t+ 2s+ 1= 0 and s+ 2t+ 1= 0. From the first equation, t= -2s- 1 so the second equation becomes s+ 2(-2s- 1)+ 1= s- 4s- 2+ 1= -3s-1= 0 and s= -1/3. In that case, t= -2(-1/3)- 1= 2/3- 1= -1/3.

The critical points are s= -1, t= -1; s= 1, t= -1; s= -1, t= 1; and s= -1/3, t= -1/3.

s= t= 1 is NOT a critical point: (1+ 1)(1+ 2+ 1)= 2(4)= 8 not 0

THanks a lot!

And sorry for the confusion, I meant dz/ds and dz/dt indeed. On a side note, what's the difference between writing dz/dt and t'? just curious

 

[tiny-tim]

On a side note, what's the difference between writing dz/dt and t'? just curious

dz/dt and z' are the same

t' makes no sense (unless you mean t' = dt/dt = 1)