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# maximum watch

1. hello

this is what i have attempted so far. i can't calculate the rest? someone who can help?
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2. (Original post by SikoSiko)
hello

this is what i have attempted so far. i can't calculate the rest? someone who can help?
You've shown that there's a minimum in the interior of the square [0,2] x [0,2], so where is the maximum hiding? It must be on the boundary! So now look at the values of f(x,y) restricted to the boundary of [0,2] x [0,2]. In other words, have a look at f(0, y), f(2, y), f(x, 0) and f(x,2).
3. (Original post by SikoSiko)
hello

this is what i have attempted so far. i can't calculate the rest? someone who can help?
look at this link, page 53 onwards

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