Easy to guess, tough to prove!

$Parallelogram\quad with\quad maximum\quad area\quad is\quad a\quad rectangle.\\Let\quad the\quad lenght\quad and\quad breadth\quad be\quad l\quad and\quad b.$ $l+b=120\\ \frac { l+b }{ 2 } \ge \sqrt { lb } \\ therefore,\quad \sqrt { lb } \le 60\\ \Rightarrow lb(maximum)=3600$

Moreover, this parallelogram will be a square.

A parallelogram with side lengths $a,b$ and one of its angles $\theta$ has area $ab \sin \theta$ . This is easy to observe by cutting the parallelogram with one of the diagonals, resulting two congruent triangles with two sides having length $a,b$ and the angle between them being $\theta$ (or $\pi - \theta$ ). Thus, fixing $a,b$ , the parallelogram with largest area must have $\sin \theta = 1$ or $\theta = \frac{\pi}{2}$ . Thus our parallelogram must be a rectangle. It's a well-known problem that among rectangles of fixed perimeter, the one with the largest area is a square. Thus the sought shape is a square with side length $60$ , and thus area $\boxed{3600}$ .

It's fairly easy to prove, if you have Calculus. (For those wondering, I tutor students in maths ranging from basic arithmetic through the calculus. Working problems at a variety of levels helps me stay in a good "zone". Additionally, this is the exact topic I am taking in my current calculus class.).

The area of a parallelogram is given by the magnitude of the cross product of vectors that form two adjacent sides.

This is given by |U x V|=|U||V|sin(t), where t is the angle between U and V. We are given that the figure is to be a parallelogram, with perimeter 240. So we know that the lengths of two adjacent sides will be 120. Let the side corresponding to U be length (magnitude) s, so the magnitude of V is 120-s.

This gives an area of s * (120 - s) * sin(t). First, we will find the angle t which maximizes the area. To do this, take a partial derivative with respect to t: d/dt s(120-s)*sin(t) = (s(120 - s)) cos(t). Setting this equal to zero, (s(120-s)) cos(t) = 0

Then either A: s = 0, which does not create a parallelogram

B: 120 - s = 0, or s = 120, which also does not create a parallelogram, or

C: cos(t) = 0, or t = 90 degrees. It follows that the parallelogram in question must be a rectangle.

To maximize the area of a rectangle, we first multiply length and width, as usual: A = s * (120-s) = 120s - s^2. Then we take the derivative of A: dA/ds = 120 - 2s, and set that equal to 0:

120 - 2s = 0, and s = 60. The second side is 120 - 60 = 60.

60*60 = 3600.

QED.

Let the sides of the parallelogram be equal to $a$ and $b$ . We know that diagonal divides it into 2 triangles of equal areas. Area of one triangle is equal to $\frac{1}{2}ab\sin \theta$ .therefore area of parallelogram is $ab\sin \theta$ which will be maximum when $\theta=90$ . Which implies it is a rectangle .

$a+b=120\\ \frac { a+b }{ 2 } \ge \sqrt { ab } \\ \quad \sqrt { ab } \le 60\\ \Rightarrow ab(max)=3600$ Which is the solution.

Let the sides be l and b

l + b >= 2 root(lb)

3600 >= lb

you might be wondering why i choose area to be l.b because square has largest area among parallelogram