# Find the size of the central angle of a sector with the largest area

Question: The perimeter of a central sector is equal to the sum of two radii and the length of the arc subtending the central angle of the sector. Among all sectors with a fixed perimeter that can be cut out of circles with different radii, find the size of the central angle of the sector with the largest area.

I am very confused, because wouldn't the largest sector just be the whole circle itself? However, the correct answer is 2 rad.

• The various sectors that you’re comparing don’t have the same radii.
– amd
May 21, 2017 at 5:09

For the given perimeter $P$, radius $r$ and arc $a$ are related as:

$$P=2r+a$$

The area of that sector is:

$$A=ar/2 \\ A(r)=(P-2r)r/2=-r^2+Pr/2$$

Hence the maximum area for the function $A(r)$ is at the center of the quadratic function, because the second derivative is negative (sign of $r^2$)

$$dA/dr=-2r+P/2=0\\ r=P/4$$

Hence the value of $r$ is found and the arc $a$ is: $$P=2P/4+a\\ a=P/2=2r$$

From my understanding of your question, you wish to solve the following optimisation problem: $$\text{Given} \ \ 2r+r\theta=C \ \ \text{for constant C, find }\arg_\theta\max \frac{r^2\theta}{2}$$

If this is what you are looking for, then it can be solved with the Cauchy-Schwarz Inequality: $$C=2r+r\theta\geq2\sqrt{2r^2\theta}\implies \frac{r^2\theta}{2}\leq\frac{C^2}{16}$$ and equality is held when $2r=r\theta$, or equivalently, $\theta = 2$.