A farmer has 30yd of fencing and wants to enclose an area beside his barn. What are the dimensions of the region of maximal area, and what is that maximal area if the region is an is isosceles triangle with its base along side of the barn. Given -> legs of length 15, x = 1/2 base

Answer 1

let the base be 2x ,(if x = 1/2 the base)
and let each of the equal sides be y
so we know 2x + 2y = 30
x+y = 15, or y = 15-x

let the height of the triangle be h
then h^2 + x2 = y^2
h^2 + x^2 = (15-x)^2
h^2 + x^2 = 225 - 30x + x^2
h = √(225-30x)

area = (1/2)base x height
= (1/2)(2x)(√(225-30x) )
= x (225-30x)^(1/2)

d(area) = x(1/2)(225-30x)^(-1/2) (-30) + (225-30x)^(1/2)
=0 for a max of area

(225-30x)^(1/2) = 15x/(225-30x)^(1/2)
cross-multiply
15x = 225-30x
45x = 225
x = 5

then h = √(225-150) = √75
and since x+y = 15
y = 10
so each side of the triangle must be 10 yds
and the largest area is (1/2)(10)(√75)
= 25√3

Makes sense that the triangle of largest area would be an equilateral triangle, just like the largest quad would be a square, etc.