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I would like to know how we can arrive at the 'l' and 'b' of the rectangle with largest area that can be inscribed in a circle using calculus. The reason why I'd want an answer using calculus is because I've seen how derivative of (area equation) is equated to ZERO to suggest "maximizing the area" - (I would like to understand that better since from what I know derivative (of an equation) gives the slope at that point for a curve).

Looking for insight into how derivative of the area equation maximizes the area.

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Without loss of generality, you can assume that your circle of radius $r$ is centered in $(0,0)$. In that case, you can parametrize your rectangle in terms of the angle $\theta$ between a diagonal and the $x$-axis, as in the picture below:

Rectangle

Then, as shown above, the area of the shaded triangle is $$A_T(\theta) = \frac{1}{2} \big(r\cos(\theta)\big) \big(r\sin(\theta)\big)$$ and, hence, the area of the rectangle is $$A_R(\theta) = 8\cdot A_T(\theta) = 4 \big(r\cos(\theta)\big) \big(r\sin(\theta)\big) = 2r^2\sin(2\theta).$$

Finally, to maximize it, as you mentioned, you just need to look for the $\theta$ such that $\frac{dA_R}{d\theta}(\theta)=0.$

Note that $A(\theta)$ gives you the area of the rectangle in terms of $\theta$. The derivative $\frac{dA_R}{d\theta}$ expresses the rate at which the area increases or decreases when variating $\theta$. So, if $\frac{dA_R}{d\theta}=0$ for some $\theta$, it means that you are in a minimium or a maximun, because there is no variation at that point: Minmax

It is like climbing a mountain: you know you are on the top when there slope there is no slope remaining to climb.

Once you know where the critical point is, you will need to determine whether it is a minimum or a maximum. For that you have to check the sign of the second derivative: if $\frac{dA_R}{d\theta}(\theta)>0$ for that value of $\theta$, then it is a minimum; if $\frac{dA_R}{d\theta}(\theta)>0$, then it is a maximum.

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    I would like to know why derivative is applied to the area equation to maximize. I'm seeking for the reason/intuition behind that. – 2017-02-13
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    @Komal-SkyNET I added some more clarifications. Have a look at the edit. – 2017-02-13
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    Thanks for that. That really helped. But I'm still confused on this part - when we find the derivative to the area function and equate it to zero, can that also not mean minimizing? In this case, the derivative also can signify a rectangle that's basically a line right? How do I know minimizing the area is equating derivative to 1 and maximizing is equating derivative to 0. – 2017-02-13
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    @Komal-SkyNET Check the final comment on the answer. For optimizing something (either maximizing or minimizing) you need to look for values that make 0 the first derivative. The sign of the second derivative on those values will determine if they are maximums or minimums. – 2017-02-13
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    Thanks, got it. Could you please point me to some documents/links so that I can understand the proof in detail? – 2017-02-13
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    @Komal-SkyNET You can find plenty of notes on this topic in Google. But, for instance you can find something simple and nice [here](https://www.mathsisfun.com/calculus/maxima-minima.html). By the way, if you found the answer useful, maybe you could accept it or upvote it so that other users know that they can use it too. – 2017-02-13
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    Sure. I tried up-voting it right after the "mountain analogy" edit, but apparently I'm newbie here...so looks like I gotta wait. But thanks a lot for your help! – 2017-02-13
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ANY function attains max/min when derivative is zero. Graphically at that stage the slope/gradient is zero, curve tangent is parallel to x-axis.

Physically it means that when you are temporarily neither increasing nor decreasing you are at the topmost or bottommost value of the dependent variable.

There is a lot of difference between a derivative being considered at all and derivative being considered as being equal to zero.

$$ x^2+y^2 = 1 \tag1$$

$$ A= 4 xy \tag2$$

$$ A = 4 x \sqrt {1-x^2} \tag3 $$

$$ \frac{dA}{dx}{\boxed{=0 }} \rightarrow x=y. \tag4$$

EDIT1:

To verify if it is a maximum the second derivative should check negative. For minimum it should be positive. It first and second derivatives are both zero then the tangent is horizontal, however it is neither max or min but is a point of inflection, i.e., locally straight.

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    Derivative zero does *not* imply a maximum or minimum: consider $f(x) = x^3$ at $x = 0$. The converse is true of a differentiable function. – 2017-02-13
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    Yes indeed,thanks.. extended it. – 2017-02-13