Question 1019175
{{{x}}}= length of one of the shorter sides (legs) of the right triangle.
{{{16-x}}}= length of the other one of the shorter sides (legs) of the right triangle.
Because the shorter sides (legs) of the right triangle are perpendicular to each other,
we can consider the length of one to be the base of the triangle,
and the length of the other one to be the height.
So, the area is
{{{area=x(16-x)/2}}}<-->{{{area=-x^2+16x}}}
Completing the square:
{{{area=(-x^2+16x)/2}}}
{{{area=(-x^2+16x-64+64)/2}}}
{{{area=(-x^2+16x-64)/2+32}}}
{{{area=(-1/2)(x^2-16x+64)+32}}}
{{{area=(-1/2)(x+8)^2+32}}}
Since the first term is negative or zero,
{{{(-1/2)(x+8)^2<=0}}} ,
{{{area<=32}}} .
In other words, the largest area possible for that right triangle is {{{highlight(32)}}} .