Question

Find the angle that maximizes the area of the isosceles triangle whose legs have length l=41

Answer 1

I'm looking at the formula for area of any triangle as A = one-half the product of any two sides and the sine of the included angle.

A = (1/2) (41)(41) sin (theta)

To maximize area, what value of theta gives sin(theta) its maximum value. That's the answer.

Answer 2

The sine function varies between -1 and 1 so what angle makes sine = 1.

Answer 3

area = 0.5 * 41^2 * sin x

differentiate and equate to zero
solve to get maximum area angle x

Answer 4

If I diff. the area formula A=SIn(x)*(1/2)41^2 and set equal to zero x=pi

Answer 5

I'm on WebWork hw site. Tried that as the answer but returned it as incorrect

Answer 6

x = ± pi/2

Answer 7

yup - required angle in degrees is 90

Answer 8

my fault it is pi/2.

Answer 9

as directrix said sine pi/2 = 1

Answer 10

I know...*brain fart*