Asked by saud

Find the angle that maximizes the area of the isosceles triangle whose legs have length l=31


angle=
Answered by Reiny
let the angle at the top be Ø
then
area = (1/2)(31)(31)sinØ
= 961/2 sinØ
d(area)/dØ = -961cosØ = 0 for a max of area
cosØ = 0
Ø = 90°
Answered by saud
it says the answer is incorrect
Answered by Reiny
Perhaps they wanted the acute angle, which of course would have to be 45° each.
I would assume you might have tried that.

If not, then they are wrong!
Answered by Jonathan
They want the answer in radians not degrees. So it's pi/2
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