Question

find the abs max and min values of f on the set D
f(x,y) = 4x+6y -x^2-y^2; D ={ (x,y|0<=x<=4,0<=y<=5 anyone guides me ,please

Answer 1

How would you like to do it?

Completing the square in x and y seems like fun. Did you have something else in mind?

Answer 2

I have many things in mind, just steps. let me show you mine and I need you check whether mine is wrong or not. my pro's way is too unclear. so I confused every thing. give me awhile

Answer 3

someone check mine, please. I do it in bewildered feeling.

Answer 4

@Hoa you are correct till (x,y)=(2,3)
you got one critical point. now, you have to check if f is at a maxima or a minima there.
conditions:
\[
\text{find and evaluate: } f_{xx},f_{yy},f_{xy}\;\text{at that point}\\
D=f_{xx}\cdot f_{yy}-f_{xy}^2\\
\text{1. }D>0 \;\&\;f_{xx}>0\implies\text{relative minima}\\
\text{2. }D>0 \;\&\;f_{xx}<0\implies\text{relative maxim}\\
\text{3. }D<0 \implies\text{stationary point}\\
\text{4. }D=0 \implies\text{test failed!}\\
\]

Answer 5

also, once you set \[f_x=0\quad\&\quad f_y=0\]
the two equations must be satisfied simultaneously.. for future reference

Answer 6

so, if the question just ask about the abs max, min without any boundary , we must follow your step. and if it asks about abs max, min with a boundary set like " a triangular vertices of (0,5),(2,6),(5,3)" for example, then, follow my steps?

Answer 7

I have 2 kinds of this topic, both ask to find max, min, local and abs, one has boundary set, one has not. and another one is Lagrange Method. confused!!!

Answer 8

your steps are the same...
but for this problem, there is only one critical point.. (2,3)

Answer 9

:(((!!! you mean it has just only one critical point? how about the ending points of the bounded set?