Question

Hello everyone!

Could you please help me with this one, I do not even know where to begin D:

Show that of all isosceles triangles with two equal sides L and L, the one with the largest area is the one whose two equal sides are perpendicular.

The only thing I can think of is plugging numbers in, but I know it probably has smth to do with optimization??

Thank you so much in advance!

Answer 1

let the angle between the two equal sides be Ø

area = (1/2)(L)(L)sinØ , (you should know that formula)

d(area)/dØ = (1/2)L^2 (-cosØ) , where L is a constant
for a max of area
(1/2)L^2(-cosØ) = 0
cosØ = 0
but of course cos 90° =0
thus Ø = 90°

and for a maximum area the two equal sides must be perpendicular to each other