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Q.How do we find out the sum of absolute maximum and absolute minimum values of $f(x,y)=(x+y)^2-(x+y)+1$ on a unit square $\{(x,y):0

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write your $$f(x,y)$$ in the form $$(x+y)^2-2\cdot \frac{1}{2}(x+y)+\frac{1}{4}+\frac{3}{4}$$ and this is $$f(x,y)=\left(x+y-\frac{1}{2}\right)^2+\frac{3}{4}$$

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    However, the original question is very poorly posed. The given function on the given domain doesn't have a maximum! It has a supremum, but it doesn't attain its supremum value. It would've been $f(1,1)$, but the boundary of the square is not included in the given domain. – 2017-01-21
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    You solved the problem! that difference does not exist. Anyway, I suggest in order to calculate were the supremum is a variable change $s=x-y\;t=x+y$. – 2017-01-22
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    After this tried but still taking partial derivative w.r.t both x and y gives the same value . can you tell me the way for ward . I am still stuck with the rt- s^2 =0 condition. – 2017-01-22