Question

No Question

Answer 1

No Answer :D

Answer 2

Then why did you ask no questions.

Answer 3

@cambrige
You can close this now. Thank you.

Answer 4

dude,really??

Answer 5

i was checking my equations

Answer 6

\[$A$=Area of isosceles $\Delta APP'$ $\;\;=\large\frac{1}{2}$$\times PP'\times AM$ $\;\;=\large\frac{1}{2}$$ab(2b\sin \theta)(a-a\cos\theta)$ $\;\;=ab[\sin\theta-\large\frac{1}{2}$$\sin 2\theta]$ Step 2: Differentiating with respect \to $\theta$ $\large\frac{dA}{d\theta}$$=ab(\cos\theta-\large\frac{1}{2}$$\cos 2\theta.2)$ $\quad\;=ab[\cos\theta-\cos 2\theta]$ For maxima and minima $\large\frac{dA}{d\theta}$$=0$ $ab[\cos\theta-\cos 2\theta]=0$ $\cos 2\theta=\cos\theta$ $\theta=\large\frac{2\pi}{3}$ Step 3: On double differentiation we get, Now,$\large\frac{d^2A}{d\theta}$$=ab[-\sin\large\frac{2\pi}{3}$$+2\sin\large\frac{4\pi}{3}]$ $\qquad\quad\;\;=ab[-\big(\large\frac{\sqrt 3}{2}\big)$$+2\big(\large\frac{-\sqrt 3}{2}\big)]$ $\qquad\quad\;\;=ab[-\big(\large\frac{\sqrt 3}{2}\big)$$-\big(2\large\frac{\sqrt 3}{2}\big)]$ $\qquad\quad\;\;=ab\big[\large\frac{-3\sqrt{3}}{2}\big]$$<0$ Step 4: $A$ is maximum when $\theta=\large\frac{2\pi}{3}$$=120^\circ$ Maximum value of $A=ab(\sin 120^\circ-$$\large\frac{1}{2}$$\sin 240^\circ)$ $\qquad\qquad\qquad\quad\;\;=ab[\large\frac{\sqrt 3}{2}-\frac{1}{2}\big(\large\frac{-\sqrt 3}{2}\big)]$ $\qquad\qquad\qquad\quad\;\;=\large\frac{3\sqrt 3}{4}$$ab$\]

Answer 7

What do the dollar signs mean.

Answer 8

nothing

Answer 9

my java script didnt work suddenly

Answer 10

in another website so copied the whole content here since this was open

Answer 11

my bad