Course:MATH102/Question Challenge/2010 December Q7
Questions? Click here to add or read comments for this problem | |
Please rate how easy you found this problem:
{{#w4grb_rate:}} Hard Easy |
---|
Question
A sector of a circle with radius and opening angle has area . Find and for the sector with the smallest perimeter, given that the area . Note that the perimeter consists of two radii and a circular arc. Be sure to verify that your answer is the global minimum.
Hints
Hint 1 |
---|
What is the length of the part of circle perimeter subtended by the angle ?. Note that if then we have the entire circular perimeter, which is |
Hint 2 |
---|
Use the fact that the area is constant to eliminate one of the variables from the equation for the perimeter. |
Solutions
Solution |
---|
We are told that the area of the shape is constant and that: . 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 : . We want to minimize . We first manipulate the constraint to isolate
The next step is to substitute in this value of into the equation for the perimeter: , , . This is a function of a single variable, , so now, to find a critical point, set and solve for the radius:
. Rejecting the negative root, we substitute into the equation for : radians To check that these values of and produce a sector with minimal perimeter for a fixed area of 9, we determine the sign of the second derivative . For any positive value of , we see that , so the critical point is a local minimum. (It is optional to actually plug in and find that . The exact value does not matter, as we already know that it is positive.) We conclude that the values we have found, namely units, radians produce a sector of minimal perimeter. |