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Minimum value of $$\bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$ for real $x\geq 0,y\in[-2,2]$

Try: Using Partial derivative

$$f(x,y) = \bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$

$$\frac{df(x,y)}{dx}=2\bigg(x-4-\sqrt{4-y^2}\bigg)+\frac{4}{\sqrt{x}}\bigg(4\sqrt{x}-y\bigg)$$

$$\frac{df(x,y)}{dy}=-2y\frac{\bigg(x-4-\sqrt{4-y^2}\bigg)}{\sqrt{4-y^2}}-2\bigg(4\sqrt{x}-y\bigg)$$

Put $\displaystyle \frac{df}{dx} =0$ and $\displaystyle \frac{df}{dy}=0$

But these $2$ equation is tedious work

Could some help me how to solve it , Thanks in advance

Ross Millikan
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DXT
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2 Answers2

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For $0\leq x\leq 4$ we have

$$f(x,y)-f(x,2)=2\left(4\sqrt{x}(2-y)+(4-x)\sqrt{4-y^2}\right)\geq 0 $$

$$f(x,y)\geq f(x,2)\geq \underset{0\leq x\leq 4}{\min }f(x,2)= \underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$$

For $x\geq 4$, use the Cauchy–Schwarz inequality, to show $f(x,y) \geq 36$ with $f(4,2)=36$ as follows

$$\left(x-4,4\sqrt{x}\right).\left(\sqrt{4-y^2},y\right)\leq 2(x+4)$$

$$f(x,y)-36=-16+8 x+x^2-2\left((x-4) \sqrt{4-y^2}+4 \sqrt{x} y\right)$$

$$\geq -16+8 x+x^2-4(x+4)=(x-4)(x+8)\geq 0$$

EDIT:

$\underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$ occurs at the zero of the cubic polynomial $x^3+8x^2+16x-16$ with $x \approx 0.7186$ and the minimum value $\approx 12.7019$

Lozenges
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I actually get another answer from @Lozenges

If we look at the given expression carefully, we will notice that it is kind of like a distance formula.

Set $ \vec{x}=(x-4,4\sqrt{x})$ and $\vec{y}=(\sqrt{4-y^2},y)$ , the given expression can be written as $|| \vec{x}-\vec{y}||^2$. The point $\vec{x}$ has a locus of a parabola $y^2 =16x+64, x\ge0, y\ge0$ and the point $\vec{y}$ has a locus of a circle$x^2+y^2=4, x\ge0$. Draw their graph and you will see it is very trivial that the closest distance between the two graphs is 6, hence the minimum value of the given expression is 36.

The idea should be correct, but I might have made a mistake somewhere? Comments are more than welcome.

Anson NG
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  • yes, that is true for the part of the parabola in the first quadrant. How about points in the second quadrant such as $(-4,0),(-3,4)$ ? These points are "closer" and therefore will give smaller values of $f(x,y)$ e.g. $f(1,2)=13$ which is less than $36$ – Lozenges Feb 24 '19 at 18:05
  • @AnsonNG your idea is right and nice. But I think you assumed wrongly for parabola $x \geq 0$ while that has to be $x-4 \geq 0.$ There is a similar mistake in the case of the circle. $x$ and $x$ is not the same in the expression of $\vec{x}$ or $\vec{y}$ and in the equation of parabola or circle. – user376343 Feb 28 '19 at 11:06