Using calculus, finding 'l' & 'b' of the rectangle with largest area that can be inscribed in a circle

Question

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.

Answer 1

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:

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:

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.

Answer 2

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.