A calculus problem by Akhil Bansal

Let $x$ be the angle between the lower base of the trapezium and each of the angled sides. Then the area $A$ of the trapezium will be that of two triangles with height $a\sin(x)$ and base $a\cos(x)$ plus that of a rectangle with width $a$ and height $a\sin(x).$ We thus have that

$A(x) = 2*\dfrac{a^{2}\sin(x)\cos(x)}{2} + a^{2}\sin(x) = a^{2}\sin(x)(1 + \cos(x)).$

To find any critical points we must find where $\dfrac{dA}{dx} = 0.$ Using the product rule we find that

$\dfrac{dA}{dx} = a^{2}(\cos(x)(1 + \cos(x)) + \sin(x)(-\sin(x))) = a^{2}(\cos(x) + \cos^{2}(x) - \sin^{2}(x)) = 0$ when

$\cos(x) + \cos^{2}(x) = \sin^{2}(x) = 1 - \cos^{2}(x)$

$\Longrightarrow 2\cos^{2}(x) + \cos(x) - 1 = 0 \Longrightarrow (2\cos(x) - 1)(\cos(x) + 1) = 0.$

Clearly we are looking for $x \in (0, \frac{\pi}{2}),$ and so the only relevant solution is $\cos(x) = \dfrac{1}{2} \Longrightarrow x = \dfrac{\pi}{3}.$ Since $A(0) = 0$ and $A(\frac{\pi}{3}) \gt 0$ we can conclude that this critical point yields a maximum for $A,$ which comes out to

$A_{max} = a^{2}\sin\left(\dfrac{\pi}{3}\right)\left(1 + \cos\left(\dfrac{\pi}{3}\right)\right) = a^{2}*\dfrac{\sqrt{3}}{2}\left(1 + \dfrac{1}{2}\right) = \boxed{\dfrac{3\sqrt{3}}{4}a^{2}}.$

I used the formula of Brahmagupta. Place with x side, the area is given by F (x) = A² = 1/4 (a + x) ³ (3a-x) the derivative is: F '(x) = 1/4 (a + x) ² (8a-4x) It has a maximum at x = 2a then just replace

Hey come on is it really essential.....................I am Brian sir had worked a whole lot on providing such a Great solution but i do not see any need of it.......And neither i think so this ques. is from Calculus Akhil please change the topic to GEOMETRY :) One just requires elementary knowledge of geometry to solve this :)