Let $x$ and $y$ be real numbers. Find the smallest possible value of $4x^2+(x+2y-6)^2+16y-23$. What method should I use?

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  • $\begingroup$ To complete the squares is always an option. $\endgroup$ – A.Γ. May 3 '17 at 14:43
  • $\begingroup$ Possible hint: note that $4x^2$ and $(x+2y-6)^2$ are always $\geq 0$. $\endgroup$ – Zain Patel May 3 '17 at 14:44
  • $\begingroup$ How to do it? It sounds easy but it is actually difficult? Is there any other method such as AM-GM inequality or Jensen's inequality? $\endgroup$ – Ray Cheng May 3 '17 at 14:44
  • $\begingroup$ Take the first partials to find all the stationary points. Then verify which are actual mins through the second partials. $\endgroup$ – msitt May 3 '17 at 14:46
  • $\begingroup$ I know how to take partial derivative but what's next? I only know about stationary point in scalar function but not vector-valued function. $\endgroup$ – Ray Cheng May 3 '17 at 14:47

Call this function $f$. If a minimum $\mathbf{x}$ exists, then it should satisfy $$\nabla f(\mathbf{x}) = 0$$ Calculating this gives us $$f_x = 10x+4y-12=0$$ $$f_y = 8y+4x-8=0$$ Or, simplified, $$5x+2y=6$$ $$x+2y=2$$ So $x=1,y=1/2$, and the minimum value is $f(1,1/2)=5$.

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  • $\begingroup$ The gradient is zero, that "zero" is zero vector? $\endgroup$ – Ray Cheng May 3 '17 at 14:59
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    $\begingroup$ you have yet to prove that it is minumum not maximum. $\endgroup$ – Leox May 3 '17 at 15:04
  • $\begingroup$ you mean take the second partial derivative? $\endgroup$ – Ray Cheng May 3 '17 at 15:08

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