Maximum Area Problem

In the figure, $AB$ has the same length as $CD$ , thus $6x = 8y$ .

Now, the area of the rectangle is given by $wl = (6x)(10 - (8x + 6y)) = (6x)(10 - (8x + \frac{9x}{2})) = 60x - 75x^2.$ . This is a quadratic equation which means it attains its maximum value at $x = -\frac{b}{2a} = -\frac{60}{2(-75)} = \frac{2}{5}.$ Plugging this back into the area equation gives a maximum area of 12.

We can also solve this using trignometry

Let CF = h , EF =a , FG =b , CD =x and BC=y

From right triangle EFG

$sin\angle EGF\quad =\quad \frac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }$

and

$cos\angle EGF\quad =\quad \frac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }$

In right triangle CDF

$cos\angle DCF\quad =\quad \frac { h }{ x }$

$\frac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \quad =\quad \frac { h }{ x }$ $(\because \quad \angle DCF\quad =\quad \angle EGF)$

$x\quad =\quad \frac { h\sqrt { { a }^{ 2 }+{ b }^{ 2 } } }{ b }$

In right triangle BCG

$CG\quad =\quad b\quad -\quad h$

$\therefore \quad sin\angle CGB\quad =\quad \frac { y }{ b\quad -\quad h }$

$\frac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \quad =\quad \frac { y }{ b\quad -\quad h }$ $(\because \quad \angle CGB\quad =\quad \angle EGF)$

$y\quad =\quad \frac { a(b\quad -\quad h) }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }$

Therefore the area of rectangle ABCD is

$xy\quad =\quad ah-\frac { a{ h }^{ 2 } }{ b }$

Which is a quadratic expression like

$Bx\quad +\quad A{ x }^{ 2 }$

whose maximum value is at

$x\quad =\frac { -B }{ 2A }$

Here which is

$h\quad =\quad \frac { b }{ 2 }$

Therefore the maximum area is

$\frac { ab }{ 4 }$

Put a and b to get answer