Maximize the inradius given the base and the area of the triangle

Question

BdMO 2013 Secondary:

A triangle has base of length 8 and area 12. What is the radius of the largest circle that can be inscribed in this triangle?

Let $A,r,s$ denote the area,inradius and semiperimeter respectively.Then we have that

$A=rs$

$\implies 12=r\dfrac{8+b+c}{2}$ [$b,c$ are the lengths of two other sides]

The only thing that can be said about $b+c$ is that it must be greater than $8$.Also,to maximize $r$,we need to minimize $b+c$.Also,$b+c$ can be rewritten as $8+2y$ for some $y$.But I am unable to infer anything else from this information.A hint will be appreciated.

Answer 1

For b+c to be minimum b=c should hold. So b=c=5.(by simple Pythagoras). So r=1.34.

Answer 2

Not entirely sure if this is correct but I think you can equalise the sums from the following:

Since you know the hight of the triangle is 3, you can write a general expression for length of the two sides $b$ and $c$ in terms of $x$. $$ b^2=9+(x)^2 \;\; \text{ and } \;\; c^2=9+(8-x)^2 $$ Minimising the sum $b+c$ is the same to minimising the sum of the squares so adding the two above equations then differentiating gives: $$ (b^2+c^2)'=2 x-2 (8-x)$$ which after solving for a minimum gives $x=4$ meaning that the perpendicular must be the in the centre of the base hence it much be an isosceles triangle thus $b=c$.

Like I said not sure if this is a valid way to show it, just seemed to work.