Question

Find the absolute extrema of f(x,y)=xy-2x bounded by the triangular region R with vertices (0,0), (0,4), (4,0).

Answer 1

Like the answer to your other question, you have a triangular region here. Examine first the partial derivatives of f and see if you can find an extrema.

Then look along the boundary. Here, you need to look at it in three pieces for the three sides.

For example, the side between (0,0) and (4,0) is given by \( (x,y) = (t,0), 0 \leq t \leq 4 \) and on that piece
\[ f(x,y) = f(t,0) = -2t \]

Answer 2

For this item, I actually have answers, but when I look at the graph of \(f(x,y)=xy-2x\), they seem wrong. It's assumable that said function is a plane, correct? My answers indicate that the maximum is at (1,0), which does seem off. \[\]

Answer 3

(1,0) is clearly not right.

Answer 4

Ah, my bad. The coordinates I got are supposed to be (1,3). But that doesn't change things much, does it?

Answer 5

grad f = (y-2,x) = (0,0) iff (x,y) = (0,2).

But that point isn't in the interior of the region. So now you just need to look at the boundary.

Answer 6

From (0,0) to (0,4), x=0. So we can say that \(f(0,y)=0, y\in[0,4].\) The derivative of f(0,y) is still 0, and all x does not affect the function.

From (0,0) to (4,0), y=0. \(f(x,0)=-2x, x\in[0,4]\).
Let B(x)=-2x
B'(x)=-2

B(0)=0
B(4)=-8
B(2)=-4

From (0,4) to (4,0), y=-x+4.
\(f(x,-x+4)=x(-x+4)-2x,x\in[0,4]\)
Let \(C(x)=-x^2+2x\)
\(C'(x)=-2x+2=0\) when x=1

C(0)=0
C(4)=-8
C(1)=1
\[\]