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I have to minimize this: \begin{align*} \min&\quad{ (x-3)^2+(y-1)^2} \\ s.t.& \quad 2x+y \leq 2 \\ &\quad x^2 + 2y = 3\\ &\quad x, y \geq 0 \end{align*} Can I isolate $y$ in the second constraint and substitute it in the first?

Valentina Sau
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    Yes, that's a reasonable approach. But before you do this, I'd recommend drawing a picture to help you understand the geometry of this problem. – Neal Jun 10 '19 at 16:17
  • @Neal I would say that's a circle with center (3,1) and the feasible region is the partial parabola under the line $2x+y-2$, but according to wolfram the function is a paraboloid – Valentina Sau Jun 10 '19 at 16:32

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Let $$(x-3)^2+(y-1)^2=z~~~(1)$$ Let us put $x^2=(3-2y)$ in (1), we get $$y^2-4y-6x+13-z=0.~~~(2)$$ $z$ will attain optimum value if when the line $y=2-2x$ touches the curve (2) which is a parabola. Let us but this line in (2) $$4x^2-6x+9-z=0.~~~~(3)$$ Now demand $B^2=4AC$.This gives $z=27/4,$ the answer.

Z Ahmed
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  • If we put $x^2=(3-2y)$ in (1) don't we get: $x^2+9-6x+y^2+1-2y=(3-2y)+9-6x+y^2+1-2y=y^2-4y-6x+13-z=0$? And following your calculus with the line $y=2-2x$: $(2-2x)^2-4(2-2x)-6x+13-z=4+4x^2-4x-8+8x-6x+13-z=4x^2+2x+9-z=0$ and here we have a negative $\Delta$. – Valentina Sau Jun 10 '19 at 17:04
  • @ValentinaSau I have re-checked to find that my steps are right! – Z Ahmed Jun 11 '19 at 01:58