Minimum value of algebraic expression This is a high school algebra problem,
If the minimum value of the expression $f(a,b) = \sqrt{a^2 + b^2 - 10a - 10b +50} +\sqrt{b^2 -4b +20} + \sqrt{a^2 - 14a +74}$ is $k$.
Which occurs at $a = \alpha$, $b = \beta$.
Find the value of $k + 4{\alpha} + 3\beta$
It can be seen that the expression can be written as $f(a,b) = \sqrt{(a-5)^2 + (b-5)^2} + \sqrt{(b-2)^2 + 4^2} + \sqrt{(a-7)^2 + 5^2}$
How do I proceed further from here and could there be a geometric solution?

Replace $a\to n+5$ and $b\to m+5$
Now the expression simplifies to $ \sqrt{n^2 + m^2} + \sqrt{ 4^2+(m+3)^2} + \sqrt{(n-2)^2 + 5^2}$
This is the sum of distance between the following pairs of points

*

*$(n,0)$ and $(2,5)$

*$(0,m)$ and $(n,0)$

*$(0,m)$ and $(-4,-3)$
If $m\ge0$,
we get $EA>EF$ and $FG+GD>AD$ (refer the following graph). Therefore the path through $A(0,0)$ gives the minimum sum.

If $m\le0$,
construct $AC$ the angle bisector of reflex $∠EAD$. It is easy to see that $∠EAC$ and $∠CAD$ are obtuse therefore we get $EF+FC>EC>EA$ and $CG+GD>CD>AD$ (refer the following graph). Therefore the path through $A(0,0)$ gives the minimum sum.

The minimum is obtained when $(n,m)=(0,0)$ i.e. when $a=b=5$.
