The Mathematics Education Department
EMT 725, Lisa F. Garey

Area of a Sector of a Circle

Click here for the Problem statement

Here is a common max-min problem approached from a different angle; using the A.M-G.M. inequality.

Let the given perimeter be P.

Solving for theta, we get .

Substituting this into area, we have .

Multiplting by 2/2, we now have A= (note: this is so that the inequality works)

By the A.M-G.M. inequality, area=

with equality iff P-2r=2r , P=4r. The maximum area is 1/16 of the square of its perimeter.

Hence, the area of a circular sector is maximized when the perimeter is 4 times the radius.

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