Sunday

January 25, 2015

January 25, 2015

Posted by **sridhar** on Wednesday, September 16, 2009 at 3:18pm.

- math -
**Reiny**, Wednesday, September 16, 2009 at 3:28pmYou imposed two conditions on the problem

1. the length is twice the width, and

2. the sum of length and width is 15

so clearly 2x + x = 15

x = 5

So it no longer is a problem dealing with maximimums.

the enclosure is 5 by 10 for an area of 50

(the maximum area would have been obtained by having two equal sides of 7.5 feet for an area of 56.25 feet^2, clearly larger than the 50 from above. But your condition of length = twice the width would not have been followed)

**Answer this Question**

**Related Questions**

math - You have a 1200 foot roll of fencing and a large field. You want to make...

Algebra - To make an enclosure for chickens, a rectangular area will be fenced ...

Math - Suppose a farmer has 120 feet of fencing to make a rectangular barnyard ...

Algebra - Suppose a farmer has 120 feet of fencing to make a rectangular ...

Math - Please help with this problem! Brandon wishes to fence in a rectangular ...

Math Calculus - Three sides of a fence and an existing wall form a rectangular ...

calculus - a farmer wants to make a rectangular pen from 200 feet of fencing. he...

Math - A farmer has 12000m roll of fencing. He wants to make 2 paddocks by ...

Math - A farmer has 12000m roll of fencing. He wants to make 2 paddocks by ...

Math - A farmer has 12000m roll of fencing. He wants to make 2 paddocks by ...