Find the smallest value of the following expression: $$\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$
I tried to derive the expression with respect to $x$ and $y$ and then equal the derivative to zero and find the critical points, but I could not do that. It is complicated. Is there a way?
On
Proceeding as your first idea by derivative, we have that for
$$f(x,y)=\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$
in order to minimize we can guess that $0<x<9$ and $0<y<3$ and then
$$f_x(x,y)=\frac{x}{\sqrt{x^2 +y^2}}+ \frac{x-9}{\sqrt{(x-9)^2+4}}=0 \implies y^2=\frac{4x^2}{(9-x)^2}\implies y=\frac{2x}{9-x}$$
$$f_y(x,y)=\frac{y}{\sqrt{x^2 +y^2}}+ \frac{y-3}{\sqrt{(y-3)^2+9}}=0\implies x^2=\frac{9y^2}{(3-y)^2}\implies x=\frac{3y}{3-y}$$
from which we obtain
$$(9-x)(3-y)=6$$
and then
$$y=\frac{2x}{(9-x)}\implies 3-\frac6{9-x}=\frac{2x}{(9-x)}$$
$$\implies {5x-21}{x-9}=0 \implies x=\frac{21}5$$
and
$$y=\frac{2x}{(9-x)}=\frac{\frac{42}5}{\frac{24}5}=\frac 7 4$$
therefore the solution is $(x,y)=\left(\frac{21}5,\frac 7 4\right)$.
On
An elementary school solution. No fancy stuff!
Let $A= (0,0), B=(9,-2), C= (-3,3)$ and let $D=(x,y)$, $E=(x,0)$, $F=(0,y)$. Then we are interested for which $D$ the expression $$f(D)= \color{red}{AD}+\color{green}{EB+FC}$$ takes a minimum. Since $\color{red}{AD} =\color{green}{EF}$ we have $$f(D) = \color{green}{BE+EF+FC}$$
By the triangle inequality we have $$f(D)\geq BC = 13$$ with equality iff $F$ and $E$ are on a line $BC:\;5x+12y=51$ and that is iff $D=({21\over 5},{7\over 4})$.
Use the Minkowski inequality: $$ \begin{aligned} &\sqrt{(x-9)^{2}+4}+\sqrt{x^{2}+y^{2}}+\sqrt{(y-3)^{2}+9}\\ =&\sqrt{(9-x)^{2}+2^2}+\sqrt{x^{2}+y^{2}}+\sqrt{3^2+(3-y)^{2}}\\ \geqslant&\sqrt{(9-x+x)^2+(2+y)^2}+\sqrt{3^2+(3-y)^{2}}\\ \geqslant&\sqrt{(9-x+x+3)^2+(2+y+3-y)^2}=13 \end{aligned} $$ "$=$" holds iff $(9-x,2)$ and $(x,y)$ are linearly dependent and $(9,2+y)$ and $(3,3-y)$ are linearly dependent, i.e. $x=\dfrac{21}{5},y=\dfrac{7}{4}$.