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I have to solve the following problem, and I don't know how to proceed: $$\min \{(x-a)^2 + (y-b)^2+xy\}$$ subject to $0 \leq x \leq 1, \ 0 \leq y \leq 1$.

I have tried to use Lagrangian multipliers but, I am not sure how to put the conditions $0 \leq x \leq 1, \ 0 \leq y \leq 1$ on them.

Bernard
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frank99
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  • Maybe you could try partial derivatives w.r.t. x and y and equate each to zero to get $2(x-a)+y=0$ and $2(y-b)+x=0$ and proceed from there. – NoChance Nov 10 '18 at 14:01
  • Hessian in The interior. Lagrange multipiers on each segment of The Square. Add vertex. – Tito Eliatron Nov 10 '18 at 14:02
  • @TitoEliatron can you develop the answer? – frank99 Nov 10 '18 at 14:35
  • @NoChance, Thank you, first I make the derivatives and then I cheack for the constraints. At the end I get that a anb have to accomplish, 0<=2b-a<=1 and 0<=(a-2b)/2<=1. Do you know if I can go further? – frank99 Nov 10 '18 at 14:37
  • When you solve the 2 equations, you will get someting like $x=2(a-b)$, given that x is between 0 and 1, then 0<=2a-2b<=1 and you will get a similar inequality from solving for y, finally you will be able to see the domain of a and b and reflect that on x and y. Its possible that you just need the 2 partial derivatives to find x and y, and not get into the range of a and b. – NoChance Nov 10 '18 at 14:52
  • I see, i wonder if there is a way to do it using lagrange multipliers? – frank99 Nov 11 '18 at 17:22

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