Mr.Rush owns a dairy farm that is bordered on one side by a river.He wants to enclose a rectangular pastor that is bordered on one side by the river and on the other three sides by a 500 feet of fencing and a four foot gate.What must the dimensions of the pasture be in order for Mr. rush to get the greatest amount of grazing area and what is the greatest area for grazing?
Get the closest to a square.
(500 +4)/3 = length of side
Side * side = ?
I thought that you couldn't add 4 to 500 because when you divide it gives you an extra foot on each side. You don't want 1 foot of a gate on each side.
This is why I don't know how you would do it.
Trial and error isn't really working because I'm not getting side lengths and widths that equal 500 feet. Then you add the 4 feet of fencing.
And the directions say a rectangal so wouldnt the two opposite side pairs have to have the same amount?
See my problem? I don't even know if anything I just said is right.
It doesn't give you an extra foot on each side, but it does add to the total dimensions and area.
A square is a rectangle with equal sides.
Try my method.
Two sides would be 168, and the side across from river would be 164 of fencing plus a 4 foot gate.
Okay thanks. I understand now.
To find the dimensions that will give the greatest amount of grazing area, we need to consider the perimeter of the pasture and the constraint of the fence length and gate.
Given that there are three sides of fencing and a four-foot gate, we can assume the length of fencing required to enclose the three sides is (500 - 4) feet. Let's call this length L.
To maximize the grazing area, we need to find the dimensions that will use up the maximum length of fencing while considering the constraints.
Considering that one side of the pasture is bordered by the river, we can assume that side does not require fencing. Thus, we need to find the dimensions for the remaining two sides, which would be the width and the length of the rectangular pasture.
Let's assume the width of the pasture is W and the length of the pasture is L as mentioned earlier.
So, the equation considering the perimeter constraint would be:
2W + L = 500 - 4
Now, let's rewrite the equation to solve for L:
L = (500 - 4) - 2W
L = 496 - 2W
To find the maximum grazing area, we need to maximize the product of the length and width. The area is given by the equation:
Area = Length × Width
A = L × W
A = (496 - 2W) × W
To find the maximum area, we need to find the critical points of the equation. Let's differentiate the equation and solve for when the derivative equals zero:
dA/dW = 496 - 4W = 0
496 = 4W
W = 124
So, the width of the pasture that gives the greatest area is 124 feet.
Now, substituting W = 124 into the equation for L:
L = 496 - 2W
L = 496 - 2(124)
L = 248
Therefore, the greatest grazing area is obtained when the width of the pasture is 124 feet and the length is 248 feet. The maximum grazing area is the product of length and width:
Area = L × W
Area = 248 × 124
Area = 30752 square feet.
Hence, the rectangular pasture with dimensions 124 feet by 248 feet will provide the greatest grazing area of 30752 square feet.