Friday

July 25, 2014

July 25, 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

**Related Questions**

Algebra 2 - A rancher is fencing off a rectangular area with a fixed perimeter ...

Pre-Algebra-math - Mr.jones bought 20 yards of fencing to make pen for dog pen ...

Algebra 2 - One hundred feet of fencing is available to make a rectangular dog ...

Math - A rectangular dog pen is to be constructed using a barn wall as one side ...

math - a rectangular pen is to be constructed alongside a barn using 120 feet of...

Math - A rectangular livestock pen with THREE SIDES of fencing is to be built ...

Pre-Algebra-math - Mr.Jones bought 20 yards of fencing to make a pen for his dog...

Math - If Steve has 120 feet of fence to make a rectangular kennel and he will ...

Math - A rancher has 220 feet of fencing to enclose a rectangular corral. Find ...

Data Management - A kennel is to be enclosed with a 20m of fencing. The length ...