Let $0\lt a\lt b$
(i) Show that among the triangles with base $a$ and perimeter $a + b$, the maximum area is obtained when the other two sides have equal length $b/2$.
(ii) Using the result of (i) or otherwise show that among the quadrilaterals of given perimeter the square has maximum area.
I solved the first part using the $A.M. \gt G.M$ inequality and the formula for area $$\Delta =\sqrt{s(s-x)(s-y)(s-z)}\,, \text{ where } s= \frac{x+y+z}{2}$$
substituting, $s=(a+b)/2$, $s-y=s-k$, and $s-z=s+k-b$.
How do I do the second part?
I understand that I have to partition it into two triangles with a diagonal, but I don't see how. I can also "feel" that it should be a square, due to reasons of symmetry, but I can't formally express it.