Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$

Question

Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?

Answer 1

$f(x,y)$ can be written as $$f(x,y)=3\sqrt 2\bigg(\sqrt{\bigg(x-\frac 12\bigg)^2+\bigg(0-\frac{1}{6}\bigg)^2}+\sqrt{\bigg(x-\frac{2y-1}{6}\bigg)^2+\bigg(0-\frac{2y-1}{6}\bigg)^2}\bigg)+\sqrt{4y^2-12y+10}$$

Here, let $$g(x):=\sqrt{\bigg(x-\frac 12\bigg)^2+\bigg(0-\frac{1}{6}\bigg)^2}+\sqrt{\bigg(x-\frac{2y-1}{6}\bigg)^2+\bigg(0-\frac{2y-1}{6}\bigg)^2}$$ which can be seen as the sum of the distance from $(x,0)$ to $\bigg(\dfrac 12,\dfrac 16\bigg)$ and the distance from $(x,0)$ to $\bigg(\dfrac{2y-1}{6},\dfrac{2y-1}{6}\bigg)$.

• Case 1 : If $\dfrac{2y-1}{6}\le 0$, i.e. $y\le\dfrac 12$, then then $g(x)$ is minimized when $(x,0)$ is on the line passing through $\bigg(\dfrac 12,\dfrac 16\bigg)$ and $\bigg(\dfrac{2y-1}{6},\dfrac{2y-1}{6}\bigg)$. So, solving $$\bigg(0-\frac 16\bigg)\bigg(\frac 12-\frac{2y-1}{6}\bigg)=\bigg(x-\frac 12\bigg)\bigg(\frac 16-\frac{2y-1}{6}\bigg)$$ for $x$ gives $x=\dfrac{1-2y}{6(1-y)}$. Therefore, we have $$\begin{align}f(x,y)&\ge 3\sqrt 2\ g\bigg(\dfrac{1-2y}{6(1-y)}\bigg)+\sqrt{4y^2-12y+10} \\\\&=\frac{1}{\sqrt 2}\sqrt{\frac{2y^2-6y+5}{(1-y)^2}} + \frac{1}{\sqrt 2}\sqrt{\frac{(1-2 y)^2 (2y^2 - 6 y + 5)}{(1-y)^2}} + \sqrt{4 y^2 -12 y + 10} \\\\&=\frac{1}{\sqrt 2}\cdot\frac{\sqrt{2y^2-6y+5}}{1-y}+\frac{1}{\sqrt 2}\cdot\frac{(1-2y)\sqrt{2y^2-6y+5}}{1-y}+ \sqrt{4 y^2 -12 y + 10} \\\\&=\frac{(2-2y)\sqrt{2y^2-6y+5}}{\sqrt 2\ (1-y)}+ \sqrt{4 y^2 -12 y + 10} \\\\&=2\sqrt{4 y^2 -12 y + 10} \\\\&=4\sqrt{\bigg(y-\frac 32\bigg)^2+\frac 14}\end{align}$$Therefore, we see that if $y\le \dfrac 12$, then the minimum of $f(x,y)$ is $4\sqrt{\bigg(\dfrac 12-\dfrac 32\bigg)^2+\dfrac 14}=2\sqrt 5$ which is attained when $(x,y)=\bigg(0,\dfrac 12\bigg)$.

• Case 2 : If $\dfrac{2y-1}{6}\gt 0$, i.e. $y\gt\dfrac 12$, then $g(x)$ is minimized when $(x,0)$ is on the line passing through $\bigg(\dfrac 12,\dfrac 16\bigg)$ and $\bigg(\dfrac{2y-1}{6},\color{red}-\dfrac{2y-1}{6}\bigg)$. So, solving $$\bigg(0-\frac 16\bigg)\bigg(\frac 12-\frac{2y-1}{6}\bigg)=\bigg(x-\frac 12\bigg)\bigg(\frac 16+\frac{2y-1}{6}\bigg)$$ for $x$ gives $x=\dfrac{2y-1}{3y}$. Therefore, we have $$\begin{align}f(x,y)&\ge 3\sqrt 2\ g\bigg(\dfrac{2y-1}{3y}\bigg)+\sqrt{4y^2-12y+10} \\\\&=\sqrt{\frac{y^2 - 2 y + 2}{y^2}} + \sqrt{\frac{(2 y - 1)^2 (y^2 - 2 y + 2)}{y^2}} + \sqrt{4 y^2 -12 y + 10} \\\\&=\frac{\sqrt{y^2-2y+2}}{y}+\frac{(2y-1)\sqrt{y^2-2y+2}}{y}+\sqrt{4y^2-12y+10} \\\\&=2\sqrt{y^2-2y+2}+2\sqrt{y^2-3y+\frac 52} \\\\&=2\bigg(\sqrt{(y-1)^2+(0-1)^2}+\sqrt{\bigg(y-\frac 32\bigg)^2+\bigg(0-\frac 12\bigg)^2}\bigg)\end{align}$$Here, let $$h(y):=\sqrt{(y-1)^2+(0-1)^2}+\sqrt{\bigg(y-\frac 32\bigg)^2+\bigg(0-\frac 12\bigg)^2}$$ which can be seen as the sum of the distance from $(y,0)$ to $(1,1)$ and the distance from $(y,0)$ to $\bigg(\dfrac 32,\dfrac 12\bigg)$. So, $h(y)$ is minimized when $(y,0)$ is on the line passing through $(1,1)$ and $\bigg(\dfrac 32,\color{red}-\dfrac 12\bigg)$. So, solving $$(0-1)\bigg(1-\frac 32\bigg)=(y-1)\bigg(1+\frac 12\bigg)$$ gives $y=\dfrac 43$. Therefore, we see that if $y\gt \dfrac 12$, then the minimum of $f(x,y)$ is $2h\bigg(\dfrac 43\bigg)=\sqrt{10}$ which is attained when $(x,y)=\bigg(\dfrac{5}{12},\dfrac 43\bigg)$.

In conclusion, it follows from the two cases that the minimum of $f(x,y)$ is $\color{red}{\sqrt{10}}$ which is attained when $(x,y)=\bigg(\dfrac{5}{12},\dfrac 43\bigg)$.

Answer 2

Complete the square in the first two terms and let $Y=2y-3$ and $X=3x-3/2$ to get $$f(X,Y)=\sqrt{Y^2+1}+\sqrt{2X^2+\frac12}+\sqrt{2\left(X+\frac12\right)^2+\left(Y+\frac12\right)^2-2XY+\frac74}$$ so that \begin{align}f_X&=\frac{2X}{\sqrt{2X^2+1/2}}+\frac{2X-Y+1}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\\f_Y&=\frac Y{\sqrt{Y^2+1}}+\frac{Y-X+1/2}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\end{align} which can be rewritten as \begin{align}\frac{4X^2}{2X^2+1/2}&=\frac{(2X-Y+1)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\\\frac{Y^2}{Y^2+1}&=\frac{(Y-X+1/2)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\end{align} and extensively simplified to give the symmetrical forms \begin{align}X^2(4Y^2+16Y+12)+4(Y-1)X-(Y-1)^2&=0\\Y^2(4X^2+12X+5)+4(2X-1)Y-(2X-1)^2&=0\end{align} on clearing denominators. Equating the two expressions on the left gives $$8X^2Y-6XY^2+8X^2-3Y^2-2XY-4X+3Y=0$$ which factorises to $(4X-3Y)(2XY+2X+Y-1)=0$, where the two interaction terms significantly aid this observation. Substituting $X=3Y/4$ gives the polynomial $$9Y^4+36Y^3+35Y^2-4Y-4=0$$ which has roots at $Y=-2,\pm1/3$ after using the rational root theorem. Hence $X=-3/2,\pm1/4$ and a minimum of $\sqrt{10}$ is obtained at $$(X,Y)=\left(-\frac14,-\frac13\right)\implies(x,y)=\left(\frac5{12},\frac43\right).$$ In fact, it is the global minimum as the expression $2XY+2X+Y-1=0$ is an anomaly due to squaring. The square root in the third term of $f(X,Y)$ means that the condition $f_X=0$ only holds if $X\le-3/2$ or $(X,Y)=(0,1)$ and $f_Y=0$ only holds if $-3/2\le X<0$ or $(X,Y)=(1/2,0)$.

Answer 3

We have \begin{align} f(x, y) &= \sqrt{4y^{2}- 12y+ 10}+ \sqrt{18x^{2}- 18x+ 5} \\ &\qquad + \sqrt{18x^{2}+ 4y^{2}- 12xy+ 6x- 4y+ 1} \\[6pt] &= \sqrt{\frac{ ( 2y- 6 )^{2} + ( 6y- 8 )^{2}}{10}} + \sqrt{\frac{( 6x- 5 )^{2}+ ( 12x- 5 )^{2}}{10}} \\[6pt] &\qquad + \sqrt{\frac{ ( 6x+ 2y- 1 )^{2}+ ( 12x- 6y+ 3 )^{2}}{10}}\\ &\ge \frac{6- 2y}{\sqrt{10}}+ \frac{5- 6x}{\sqrt{10}}+ \frac{6x+ 2y- 1}{\sqrt{10}} \\[10pt] &= \sqrt{10}. \end{align}

Answer 4

Since $$f(x,y)= \sqrt{(2y-3)^2+1}+\dfrac{\sqrt2}2\sqrt{(6x-3)^2+1}+\dfrac12\sqrt{(6x-4y+2)^2+(6x)^2},$$ then the substitutions $$2y-3 = \sinh s,\quad 6x-3=\sinh t,\quad s,t\in\mathbb R,\tag1$$ transform $\;f(x,y)\;$ to $$g(s,t)= \cosh s + \dfrac{\sqrt2}{2} \cosh t+\dfrac12\,r(s,t),\tag2$$ where $$r(s,t)=\sqrt{\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2}.\tag3$$

The stationary points of $\;g(s,t)\;$ correspond to solutions of the system $\;g'_s=g'_t=0,\;$ or \begin{cases} \sinh s-\dfrac1{r(s,t)}\,\cosh s\,(\sinh t-2\sinh s-1)=0\\[4pt] \dfrac{\sqrt2}2\sinh t-\dfrac1{r(s,t)}\,\cosh t\,(\sinh t-\sinh s+1)=0,\\[4pt] \end{cases}

\begin{cases} r(s,t) = \coth s\,(\sinh t-2\sinh s-1)\\[4pt] r(s,t) = \sqrt2\,\coth t\,(\sinh t-\sinh s+1),\\[4pt] \end{cases}

\begin{cases} \big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2s\\[4pt] =(\sinh t-2\sinh s-1)^2(1+\sinh^2s)\\[4pt] \big(\left(\sinh t-2\sinh s-1\right)^2+\left(\sinh t+3\right)^2\big)\sinh ^2t\\[4pt] =2(\sinh t-\sinh s+1)^2(1+\sinh^2t)\\[4pt] \coth s\,(\sinh t-2\sinh s-1)\ge0\\[4pt] \coth t\,(\sinh t-\sinh s+1)\ge0, \end{cases}

\begin{cases} \left(\sinh t+3\right)\sinh s=\sinh t-2\sinh s-1\\[4pt] \big(2\sinh^2t-4\sinh t \sinh s +4\sinh^2s+4\sinh t+4\sinh s+10\big)\sinh ^2t\\[4pt] =2(\sinh^2t-2\sinh t \sinh s+\sinh^2s+2\sinh t-2\sinh s+1)(1+\sinh^2t)\\[4pt] \sinh t+3\ge0,\\[4pt] \coth t\,(\sinh t-\sinh s+1)\ge0, \end{cases}

\begin{cases} \sinh t+3 = \dfrac{2\sinh s+4}{1-\sinh s}\\[4pt] (\sinh s+2)^2\sinh ^2t =(\sinh t-\sinh s+1)^2\\[4pt] \sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt] \sinh t\,(\sinh t-\sinh s+1)\ge0, \end{cases}

\begin{cases} \sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt] (\sinh s+2)^2(5\sinh s+1)^2 =(5\sinh s+1+(\sinh s-1)^2)^2\\[4pt] \sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt] \sinh t\,(\sinh t-\sinh s+1)\ge0, \end{cases}

\begin{cases} \sinh t = \dfrac{5\sinh s+1}{1-\sinh s}\\[4pt] (\sinh s+2)^2(5\sinh s+1)^2 =(\sin s+1)^2(\sinh s+2)^2\\[4pt] \sinh t\ge -3,\quad 1 > \sinh s \ge -2\\[4pt] \sinh t\,(\sinh t-\sinh s+1)\ge0, \end{cases}

$$\begin{pmatrix}\sinh s\\ \sinh t\\ g(s,t)\end{pmatrix} = \left\{\begin{pmatrix}-2\\ -3\\ 2\sqrt5\end{pmatrix}, \begin{pmatrix}-\frac13\\ -\frac12\\ \sqrt10 \end{pmatrix}, \begin{pmatrix}0 \\ 1\\ 4\end{pmatrix}\right\}.$$

Therefore, the least value of $\;f(x,y)\;$ is $\;\color{brown}{\mathbf{\sqrt{10}}}\;$ at $\;\color{brown}{\mathbf{(x,y)=\left(\dfrac5{12},\dfrac43\right)}}.\;$

Answer 5

Recall the well-known fact: If $g(u)$ is a convex function on $\mathbb{R}^n$, then $$g(u) \ge g(v) + \nabla g(v)^\mathsf{T}(u - v), \ \forall u, v \in \mathbb{R}^n.$$

Clearly, $f(x, y)$ is a convex function on $\mathbb{R}^2$. Thus, we have $$f(x, y) \ge f(\tfrac{5}{12}, \tfrac{4}{3}) + \tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (x - \tfrac{5}{12}) + \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (y - \tfrac{4}{3}) = \sqrt{10}.$$ Also, $f(\tfrac{5}{12}, \tfrac{4}{3}) = \sqrt{10}$.

The minimum of $f(x, y)$ is $\sqrt{10}$ achieved at $x = 5/12$ and $y = 4/3$.

Another way: Since $f(x, y)$ is a convex function on $\mathbb{R}^n$ and $\tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3}) = \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) = 0$, $f(x,y)$ achieves its minimum at $x = 5/12$ and $y = 4/3$.