Course:MATH102/Question Challenge/2010 December Q7

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Question

A sector of a circle with radius $r$ and opening angle $\theta$ has area $A=r^{2}\theta /2$. Find $r$ and $\theta$ for the sector with the smallest perimeter, given that the area $A=9$. Note that the perimeter consists of two radii and a circular arc. Be sure to verify that your answer is the global minimum.

A sector of a circle of radius $r$

Hints

[show]Hint 1
What is the length of the part of circle perimeter subtended by the angle $\theta$?. Note that if $\theta =2\pi$ then we have the entire circular perimeter, which is $2\pi r$

[show]Hint 2
Use the fact that the area is constant to eliminate one of the variables from the equation for the perimeter.

Solutions

[show]Solution

We are told that the area of the shape is constant and that: $A={\frac {r^{2}\theta }{2}}=9$. This will be a constraint in this optimization problem The perimeter of the shape is the sum of the circular arc and the two radii, so is : $P=r\theta +2r$. We want to minimize $P$. We first manipulate the constraint to isolate $\theta$ ${\frac {r^{2}\theta }{2}}=9$ $r^{2}\theta =18$ $\theta ={\frac {18}{r^{2}}}$ The next step is to substitute in this value of $\theta$ into the equation for the perimeter: $P(r)=r\left({\frac {18}{r^{2}}}\right)+2r$, $P(r)={\frac {18r}{r^{2}}}+2r$, $P(r)={\frac {18}{r}}+2r$. This is a function of a single variable, $r$, so now, to find a critical point, set $P'(r)=0$ and solve for the radius: $P'(r)={\frac {-18}{r^{2}}}+2=0$ ${\frac {18}{r^{2}}}=2$ $2r^{2}=18$ $r^{2}=9$ $r=\pm 3$. Rejecting the negative root, we substitute $r=3$ into the equation for $\theta$: $\theta ={\frac {18}{r^{2}}}={\frac {18}{3^{2}}}=2$ radians To check that these values of $r$ and $\theta$ produce a sector with minimal perimeter for a fixed area of 9, we determine the sign of the second derivative $P''(r)={\frac {36}{r^{3}}}$. For any positive value of $r$, we see that $P''(r)>0$, so the critical point is a local minimum. (It is optional to actually plug in $r=3$ and find that $P''={\frac {36}{(3)^{3}}}={\frac {36}{27}}$. The exact value does not matter, as we already know that it is positive.) We conclude that the values we have found, namely $r=3$ units, $\theta =2$ radians produce a sector of minimal perimeter.