Posted by **April** on Thursday, March 10, 2011 at 1:16pm.

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?

- Algebra and Trig -
**Reiny**, Thursday, March 10, 2011 at 1:45pm
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 = 120-2x

area = xy

= x(120-2x)

= 120x - 2x^2

If you are in Calculus .....

d(area)/dx = 120 - 4x = 0 for a max area

x = 30

y = 120-2(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(x-30)^2 + 1800

max area is 1800, when x = 30

- Algebra and Trig -
**tchrwill**, Thursday, March 10, 2011 at 4:46pm
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.

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