Question 894103
Form a right triangle, sides A and B, hypotenuse C.
Perimeter:
{{{A+B+C=72}}}
Pythagorean theorem:
{{{A^2+B^2=C^2}}}
Let's get rid of {{{C}}}
{{{C=72-A-B}}}
{{{C^2=(72-A-B)^2}}}
.
.
{{{A^2+B^2=A^2+2AB-144A+B^2-144B+5184}}}
{{{0=2AB-144A-144B+5184}}}
{{{B(2A-144)=144A-5184}}}
{{{B=(144A-5184)/(2A-144)}}}
{{{B=(72(A-36))/(A-72)}}}
Area (Use Y instead of A):
{{{Y=(1/2)AB}}}
Maximize the area. 
Make Y a function of only 1 variable.
Substitute,
{{{Y=(1/2)A*((72(A-36))/(A-72))}}}
{{{Y=(36A^2-1296A)/(A-72)}}}
Take the derivative using the quotient rule,
{{{dY/dx=((A-72)(72A-1296)-(36A^2-1296A)(1))/(A-72)^2}}}
{{{dY/dx=(36(A^2-144A+2592))/(A-72)^2}}}
Now set it equal to zero.
{{{A^2-144A+2592=0}}}
{{{A^2-144A+5184+2592=5184}}}
{{{(A-72)^2=5184-2592}}}
{{{(A-72)^2=2592}}}
{{{A-72=0 +- 36sqrt(2)}}}
{{{A=72 +- 36sqrt(2)}}}
Since {{{A}}} cannot be greater than 72, 
{{{A=72-36sqrt(2)}}}
Then,
{{{B=(72(A-36))/(A-72)}}}
{{{B=(72(36-36sqrt(2)))/(-36sqrt(2))}}}
{{{B=72-36sqrt(2)}}}
It's an isosceles right triangle.
So then,
{{{Y=(1/2)(AB)}}}
{{{Y=(1/2)(72-36sqrt(2))^2}}}
{{{Y=(1/2)(5184-2592sqrt(2)-2592sqrt(2)+2592)}}}
{{{Y=(1/2)(7776-5184sqrt(2))}}}
{{{highlight(Y=3888-2592sqrt(2))}}}