*This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com*

This is a very simple question. Let M be the angle, the arc of the sector AOB makes with the center O. Let OA = OB = radius r.

Given OA + Arc AB + OB = 20 --> 2r + rM = 20 --> M = (20-2r)/r. ..... (1)

Area A of the sector = (M/2)*r*r = 10r-(r*r) ---------- From (1)

dA/dr = 10 - 2r = 0 ----> r=5

Second derivative is negative for r =5

We get maximum area at r=5

A = 25 ----------- From (1)

Sanjay Mohan Bhatnagar, Bhopal

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