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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 $\{(x,y):0<x<1,0<y<1\}$? Using the method of $rt-s^2$ its value is zero. So now how do I search for the Max and main value?

Rafa Budría
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shadow kh
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1 Answers1

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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 $$f(x,y)=\left(x+y-\frac{1}{2}\right)^2+\frac{3}{4}$$

Dr. Sonnhard Graubner
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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. – zipirovich Jan 21 '17 at 21:18
  • You solved the problem! that difference does not exist. Anyway, I suggest in order to calculate were the supremum is a variable change $s=x-y;t=x+y$. – Rafa Budría Jan 22 '17 at 00:03
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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. – shadow kh Jan 22 '17 at 01:01