What is the meaning of a stationary value in real life?

Question

So I just solved a question saying that there is a sector with radius "$r$" and sector angle "$ \theta $" radians, the total area of the sector is "$A$" and the perimeter is $40$ $cm$. The question was: Find the value of $r$ for which there is a stationary value of $A$. I found $r$ as $10$ by rearranging the sector area formula and taking the derivative, the answer is right. However I don't understand what is happening geometrically. So my question is what happens to the shape in real life when $r$ is $10$ ?

Answer 1

Quick answer:

So Your question is not specific enough. I suggest enter Your problem for example into the free wolframalpha.com text input field. This is a distinction between cases that are separately solved.

I understand your question as being the third case, find a stationary point in a specified domain with a parameter.

In Your case the parameter is A, the total area of the sector. The perimeter is 40 cm.

Now the definitory problem comes up. What is the sector in Your question? There are three choices: a. A plain problem. b. A spherical problem in 3D. c. The dimension is not given an another parameter.

An example of sector rings in 3D: An example of sectors from top to bottom in a sphere in 3D: This is done with a floor function. Both examples of sectors are shown to transparency without the boundary planes originating from the origin of the sphere in 3D.

The sectors shown in example 2 might already be considered stationary in the floor function.

In the case of the cosine the stationary points are the minima and maxima in the region defined. That is rather similar to the example with the floor function. For values smaller than one, zero is the stationary point. Each sector has A=0 as area. For values smaller than two, one is the stationary point so the Area of the sector is surrounded from the circular line and the limiting radius lines giving $\theta 1^2$.

For shortness the dimension might be one due to only giving $\theta$!

The perimeter is $2 \pi r$. so $r=\frac{40}{2 \pi} =\frac{20}{\pi} \approx 6.3662$.

The formula for a sector is then $\frac{\theta r^2}{2}$. The derivative is $\pi r$. This is zero for $r=0$. This is bigger than zero for all other radii.

If this is interpreted for fixed $\theta$ as a parameter. $r=\sqrt{\frac{2 A}{\theta}}$. Then r is a function of $\theta$. The functional behavior is $x-1$. There is only a singularity at zero and the function goes to zero at infinity. The bound is $2 \pi$. So no stationary points at all.

Situation changes if the following terms are used:

Then You can discuss $\cos$, $\sinc$ or $x,y$. Then without limit to the polar coordinates, there are stationary points. But not at the point You calculated.

Answer 2

A stationary point for an area of a triangle means that this triangle reaches a maximum area at $r=10$ where on the interval $r \in \space]0,10[ \space$the area is increasing as the value of the derivative is positive while the interval $r \in\ ]10,20[\ $the area is decrasing as the value of derivative is negative