Find the absolute ,maximum and minimum of $\sf f(x,y)= 4 x y^2-x^2y^2-xy^3$ on the close triangular region in the xy-plane with vertices (0,0), (0,6), (6,0).

Question

owlet · Answer

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owlet · Answer

$\sf fx= 4y^2-2xy^2-y^3=0\y^2(4-2x-y)=0\ .\fy=8yx-2x^2y-3y^2x=0\xy(8-2x-3y)=0$

From that, the points I found are: (0,0,0) and (1,2,4)

owlet · Answer

Now from the boundaries


L1: x=0, 0<= y<=6, I'll get (0,0,0)
L2: y=0, 0<=x<= 6, I'll get (0,0,0) 
L3: y=-x+6
I sub it in the original function and then I evaluated its derivative and found x=-6 and x=6 as roots, since x=-6, is not in the boundary, I removed it and used x=6 instead.  When x=6, y=0...so I will have point (6,0,0)

So my absolute max will be (1,2,4) and my min are (0,0,0) and (6,0,0)...

owlet · Answer

However the right answer for the minimum is (2,4,-64)

ganeshie8 · Answer

The boundary of the region are not points.
They are the 3 sides of the triangle : 
x+y = 6
x = 0
y = 0