Algebra and Trig
posted by April .
Aldo has feet of fencing. He will use it to form three sides of a rectangular garden. The fourth side will be along a house and will not need fencing. What is the maximum area that the garden can have?

missing info
how many feet of fencing?
Suppose he has 120 feet, change the solution below to the appropriate number of feet in your question
let length of field be y, (y being parallel to house.)
let the width be x
2x + y = 120
y = 1202x
area = xy
= x(1202x)
= 120x  2x^2
If you are in Calculus .....
d(area)/dx = 120  4x = 0 for a max area
x = 30
y = 1202(30) = 60
Max area = 30(60) = 1800 ft^2
if you don't take Calculus, complete the square
area = 2(x^2  60x + 900  900 )
= 2(x30)^2 + 1800
max area is 1800, when x = 30 
Considering all rectangles with a given perimeter, one side being another straight boundry, the 3 sided
rectangle enclosing the greatest area has a length to width ratio of 2:1.