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[Kamo123]

Hello

This is what I have attempted so far. But now I'm at utter loss at how to calculate the rest Can you help? Thanks in advance.

 

[Mark44]

Hello

This is what I have attempted so far. But now I'm at utter loss at how to calculate the rest Can you help? Thanks in advance.

Your function is defined on a closed, bounded region. Check the four boundaries for the maximum value.

 

[Kamo123]

Your function is defined on a closed, bounded region. Check the four boundaries for the maximum value.

What does that mean? Could you explain more explicitly?

 

[Mark44]

What does that mean? Could you explain more explicitly?

Your function is defined on the square [0, 2] X [0, 2]. Along each of the four sides of this square your function simplifies to a single-variable function. For example, on the lower edge of the square, y = 0 and x varies from 0 to 2. So f(x, y) = f(x, 0). This is a function of x alone. Any maximum value will occur where the derivative is zero or at an endpoint of this edge.

Do something similar for each of the four edges.

 

[Ray Vickson]

Hello

This is what I have attempted so far. But now I'm at utter loss at how to calculate the rest Can you help? Thanks in advance.

You say that (1/4,1) is a local minimum. It is a lot more than that: it is the global minimum in the entire plane $\mathbb{R}^2$; and because the point (1/4,1) is feasible (satisfies all the constraints) it is the overall minimum in your constrained problem. Because (1/4,1) is the only stationary point of f(x,y), no interior point (with strict inequalities 0 < x < 2 and 0 < y < 2) can possible be a maximum, local or otherwise. Therefore, as Mark44 has suggested, you need to look along the boundary lines x = ± 2 and/or y = ± 2 in order to locate a constrained maximum.