Find the least area of a triangle in the 1st quadrant formed by the x,y axes and a line L through (3,8)?

Question

It's an optimization problem for Calculus so derivates must used. Similar to related rates.

Answer 1

12 units is the least Area of the Triangle

Answer 2

any line through (3,8) ( not parallel to y axis) is
y-8=m(x-3)
for x= 0 y= -3m+8 and for y =0 x= 3-8/m

-3m+8>0 so m <8/3 and 3-8/m>0 so (3m-8)/m >0 and
m<0 and m>8/3 so m must be <0
2A= ( -3m+8)(3-8/m)
2A´= 1/2[ -3(3-8/m)+ (-3m+8) 8/m^2) = -9+24/m-24/m+64/m^2
9m^2=64 and m =-8/3
y= 16 and x= 6
A=48