2 w + L = 120 so L = (120-2w)
A = w L = w (120-2w) = 120 w - 2w^2
y = A/2 = 60 w - w^2
maximize A/2 same as max A
w^2 - 60 w = -y
w^2 - 60 w + 900 = -y + 900
(w-30)^2 = -(A/2) + 900
vertex at w = 30
the L = 120-60 = 60
so
30, 30 and 60
A rectangular garden next to a building is to be fenced with 120 feet of fencing. The side against the building will not be fenced. What should the lengths of the other three sides be in order to assure the largest possible area?
2 answers
wow you rock thanks a bunch x