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August 1, 2014

August 1, 2014

Posted by **fifi** on Monday, September 19, 2011 at 11:34am.

- math -
**Reiny**, Monday, September 19, 2011 at 11:45amMake a sketch,

let the width of the field be x,

let the length of the field be y

then y + 2x = 20 , (we need only one length)

y = 20-2x

area = xy

= x(20-2x)

= -2x^2 + 20x

I don't know at what level of math-study you are.

If you know calculus, find

d(area)/dx = -4x + 20 = 0

x = 5 , then y = 10 for a max area of 50

If you don't know calculus, complete the square of the quadratic function

A = -2x^2 + 20x

= -2(x^2 - 10x +**25 - 25**)

= -2((x-5)^2 - 25)

= -2(x-5)^2 + 50

vertex is (5,50)

so the max area is 50 when x = 5 , as above

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