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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To complete the squares is always an option. – A.Γ. May 03 '17 at 14:43
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Possible hint: note that $4x^2$ and $(x+2y-6)^2$ are always $\geq 0$. – Zain Patel May 03 '17 at 14:44
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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? – Ray Cheng May 03 '17 at 14:44
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Take the first partials to find all the stationary points. Then verify which are actual mins through the second partials. – msitt May 03 '17 at 14:46
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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. – Ray Cheng May 03 '17 at 14:47
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@RayCheng Please show your progress so far. – msitt May 03 '17 at 14:48
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$8x+2(x+2y-6)(1+2y')+16y'=0$ with respect to x – Ray Cheng May 03 '17 at 14:51
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i have found $5$ and the equal sign holds if $$x=1,y=\frac{1}{2}$$ – Dr. Sonnhard Graubner May 03 '17 at 15:00
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What is the method if I use completing squares? – Ray Cheng May 03 '17 at 15:11
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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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