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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.

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Take any diagonal of the quadrilateral, let its base be $a$. Then by the first result, two sides on it must be the same. As I can choose any diagonal, all sides have to be the same. So this gets you to a rhombus. Now show you can increase the rhombus area for the same perimeter if it's a square. –  Macavity Mar 25 at 9:52
@Macavity ah that's brilliant. –  Sabyasachi Mar 25 at 9:59
@Sabyasachi Wasnt that obvious when you knew you had do something with the diagonal and you had the first result? –  Sawarnik Mar 25 at 10:06
@Sawarnik not really. long ass exam. gets tiring. –  Sabyasachi Mar 25 at 10:08

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