I am working on this largrange multiplyer problem:
For part A how did did find the mimimum at (sqrt(10),0)

Question

I am working on this largrange multiplyer problem:
For part A how did did find the mimimum at (sqrt(10),0)

anonymous · Answer

attachment

anonymous · Answer

its looks to me like they got that by saying that y=0, but if that is the case why did they do that

anonymous · Answer

@Zarkon , could you help me sir

anonymous · Answer

i got all the other crtical points , even the ones wiithin the constraint, but i cant figure out how they found a min at (sqrt(10),0)

phi · Answer

I get 3 equations
x+Lx-1=0
2y+Ly=0
$$x^2+y^2-10=0 $$
solving, we get x=-1 --> y= ±3
or, y=0 --> x= ±sqrt(10)
the first pair is a max (plug in to find f(x,y)) = 24
the 2nd pair give a smaller number 13-2sqrt(10), so it must be a min

anonymous · Answer

oh your right, cause we factor that second equation