A rancher has
4 comma 7004,700
feet of fencing available to enclose a rectangular area bordering a river. He wants to separate his cows and horses by dividing the enclosure into two equal areas. If no fencing is required along the river, find the length of the center partition that will yield the maximum area.
3 answers
same as the others. Give it a try. period
A rancher has
4,700 feet of fencing available to enclose a rectangular area bordering a river. He wants to separate his cows and horses by dividing the enclosure into two equal areas. If no fencing is required along the river, find the length of the center partition that will yield the maximum area.
4,700 feet of fencing available to enclose a rectangular area bordering a river. He wants to separate his cows and horses by dividing the enclosure into two equal areas. If no fencing is required along the river, find the length of the center partition that will yield the maximum area.
You don't say whether the dividing fence is to be parallel to the river or not. If not, then let x be the length of the side parallel to the river. Then you have
x+3y = 4700
so, x = 4700-3y
The area is
a = xy = (4700-3y)y = 4700y - 3y^2
The vertex of this parabola is at (2350/3,5522500/3) or (783.33,1.84*10^6)
As usual, the maximum area is when the fence is divided equally among lengths and widths:
one length of 2350
3 widths of 2350/3
Now, if the dividing fence op length x is parallel to the river, you have
2x+2y = 4700
a = x(2350-x) = 2350x-x^2
and max area is at x=2350/2 and is 1.38*10^6
x+3y = 4700
so, x = 4700-3y
The area is
a = xy = (4700-3y)y = 4700y - 3y^2
The vertex of this parabola is at (2350/3,5522500/3) or (783.33,1.84*10^6)
As usual, the maximum area is when the fence is divided equally among lengths and widths:
one length of 2350
3 widths of 2350/3
Now, if the dividing fence op length x is parallel to the river, you have
2x+2y = 4700
a = x(2350-x) = 2350x-x^2
and max area is at x=2350/2 and is 1.38*10^6
1 answer