Optimization problems!?

Question

Find the area of the largest isosceles triangle having a perimeter of 18 meters.

Answer 1

Since it's isosceles, 2 sides are equal, call them x, base side is y.

P = 18 = 2x + y, so y = 18 - 2x.

Then y/2 = 9 - x, and height h is

h² = x² - (y/2)²
h² = x² - (9 - x)²
h² = x² - 81 + 18x - x²
h² = 18x - 81
h = √(18x - 81) = 3√(2x - 9)

Then area = 1/2 base times height
A = 3(9 - x)√(2x - 9)
A = (27 - 3x)√(2x - 9)
A' = -3√(2x - 9) - (3x + 27)/√(2x - 9)

max when A' = 0

0 = -3√(2x - 9) - 3(x - 9)/√(2x - 9)
√(2x - 9) = -(x - 9)/√(2x - 9)
2x - 9 = 9 - x
3x = 18
x = 6
So triangle is equilateral, area is

A = 3(9-6)√(12-9)
A = 9√3

Answer 2

Triangle with maximum area with given perimeter is equilateral.
Therefore, the side of your triangle is 18/3 = 6 meters. The area is then 9*sqrt(3) = 15.58845727.

If you want the proof of the statement on the first line here it is.
Mark the congruent side b and angle between congruent sides as 2*fi. Then the perimeter is:
P = 2*b + 2*b*sin(fi) = 2*b*(1 + sin(fi))...........(1)
and area:
A = b*sin(fi)*b*cos(fi) = (b^2/2)*sin(2*fi)...........(2)
Taking b from (1) and plugging into (2) yields:
A = (P^2/8)*[sin(2*fi)/(1 + sin(fi))^2]...............(3)
dA/dfi = 0 yields:
cos(2*fi)*(1 + sin(fi)) = sin(2fi)*cos(fi), or
cos(2*fi) = sin(2*fi)*cos(fi) - cos(2*fi)*sin(fi) = sin(fi)
1 - 2*sin(fi)^2 = sin(fi)................(4)
The soultion of (4) for positive fi is sin(fi) = 1/2 giving fi = pi/6
or 2*fi = pi/3 = 60 degrees making the triangle equilateral.

Answer 3

come up with 2 equations
a=area
b=base length
c=other side length whatever its called

a=b*1/2*sqrt(c^2-(1/2b)^2)
b=18-2c

then graph and find derivative, see where it equals 0, plug in all values into original equation to see where area is highest