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November 24, 2014

November 24, 2014

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

- Algebra and Trig -
**Reiny**, Thursday, March 10, 2011 at 1:45pmmissing 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 4:46pmConsidering 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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