ƒ¡ is a circle with center O . A and B are points on ƒ¡ such that the sector AOB has perimeter 40 . What is the measure of �ÚAOB (in radians) when the area of the sector AOB is maximized
perimeter = 2r+rθ = 40
r = 40/(θ+2)
a = 1/2 r^2 θ = 800θ/(θ+2)^2
da/dθ = 800(θ-2)/(θ+2)^3
da/dθ = 0 when θ=2 for maximum area
Note that it doesn't really matter what p is; max area for a constant perimeter is when θ=2
To find the measure of angle AOB (in radians) when the area of the sector AOB is maximized, we need to use some calculus. Let's go through the steps:
1. Let θ be the measure of angle AOB (in radians).
2. The perimeter of the sector AOB is given by the formula P = rθ, where r is the radius of the circle.
3. In this case, the perimeter is given as 40, so we have 40 = rθ.
4. Rearranging the equation, we have θ = 40/r.
5. The area of the sector AOB is given by the formula A = 0.5r²θ, where r is the radius of the circle.
6. Substituting the value of θ from step 4, we have A = 0.5r²(40/r).
7. Simplifying the equation, we have A = 20r.
8. To maximize the area A, we need to find the critical points by taking the derivative of A with respect to r and setting it equal to zero.
9. Differentiating A with respect to r, we have dA/dr = 20.
10. Setting dA/dr equal to zero, we have 20 = 0.
11. Since this equation has no solution, it means that the area A is always increasing with respect to r.
12. Therefore, to maximize the area A, we need to maximize the radius r.
13. As the radius of the circle increases, the measure of angle AOB also increases.
14. However, the maximum measure of angle AOB is limited to 2π radians (a full circle).
15. So, the measure of angle AOB when the area of the sector AOB is maximized is 2π radians.
Note: This result is independent of the given perimeter of 40. The maximum area of the sector AOB occurs when θ is equal to 2π radians, regardless of the specific values of the perimeter and radius.