Posted by **Emily** on Monday, September 14, 2009 at 7:03pm.

I usually know how to do these types of problems, but the second variable just threw me off balance..

47. A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals. What dimensions will produce a maximum enclosed area? (the diagram is of two adjacent corrals sharing a middle fence, y, and each individual corral has length x [so the two together would be 2x for the entire length])

I hope I've explained the diagram enough...any help is greatly appreciated!! :D

- Precalculus -
**bobpursley**, Monday, September 14, 2009 at 7:10pm
Fencing needed= 3Width + 4 length, correct? I dont have your diagram

Area= 2*W*L

so 200=3w+4L so l= (200-3w)/4 putting that into area..

area= 2w(200-4w)/4

- Precalculus -
**Emily**, Monday, September 14, 2009 at 8:18pm
Ohhhhhh i totally get it now! Thanks!! :D

## Answer This Question

## Related Questions

- algebra - a rancher has 310 feet of fencing with which to enclose two ...
- math - A rancher has 296 feet of fencing to enclose two adjacent rectangular ...
- College Algebra - A rancher has 100 meters of fencing to enclose two adjacent ...
- calculus - A rancher has 1000 feet of fencing with which to enclose two adjacent...
- Precalculus - There are 4 rectangular corrals of identical dimensions along the ...
- Precalculus - The owner of a horse stable wishes to set up 4 rectangular corrals...
- Algebra - A rancher wishes to enclose a rectangular partitioned corral with 1932...
- Precalculus - A rancher wishes to fence in a rectangular plot of land and divide...
- Math - if you have 200 feet of fencing to enclose four adjacent rectangular ...
- ALGEBRA 2 urgent!! - if you have 200 feet of fencing to enclose four adjacent ...

More Related Questions