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September 30, 2014

September 30, 2014

Posted by **Rachel** on Wednesday, May 8, 2013 at 1:02am.

How do you solve this problem?

- algebra 2 -
**Reiny**, Wednesday, May 8, 2013 at 6:56amI assume you are implying that we don't need a fence along the house.

let the length be y m (the single side)

let the width be x m

so we know that 2x + y = 60

or y = 60 - 2x

Area = xy

= x(60-2x)

= -2x^2 + 60x

At this point I don't if you know Calculus or not.

If you do, then

d(Area)/dx = -4x + 60

= 0 for a max of area

-4x + 60 = 0

x = 15

then y = 30

if you don't take Calculus, we have to complete the square

area = -2(x^2 - 30x)

= -2(x^2 - 30x + 225 - 225)

= -2((x-15)^2 - 225)

= -2(x-15)^2 + 450

so the area is a maximum when x = 15

and then y = 60-30 = 30

by either method,

the width has to be 15 m, and the length has to be 30 m for a maximum area of 15(30) or 450 m^2

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