Let x be the width of the corral parallel to the barn and y be the length perpendicular to the barn. The total length of the fencing is given by:
3y + 2x = 4000
We want to maximize the total area A = xy. To do this, we first need to express either x or y in terms of the other variable. From the equation above, we can solve for x:
x = (4000 - 3y)/2
Now, substitute this expression for x into the area equation:
A = y * (4000 - 3y)/2
To maximize the area, we can find the maximum of this function with respect to y by taking the derivative of A with respect to y and setting it equal to 0:
dA/dy = (4000 - 6y)/2
Setting the derivative equal to 0:
0 = (4000 - 6y)/2
0 = 4000 - 6y
Now, solve for y:
6y = 4000
y = 2000/3 ≈ 666.67 feet
Substitute the value of y back into the expression for x:
x = (4000 - 3*666.67)/2 ≈ 1000 ft
Hence, the dimensions of the corral that maximize the enclosed area are approximately 1000 feet by 666.67 feet.
To calculate the maximum area, multiply:
A = 1000 * 666.67 ≈ 667,670 square feet
a rancher has 4000 feet of fencing for constructing a rectangular corral. one side of the corral will be formed by a barn and requires no fence. three exterior fences and 2 interior fences partition the corral into 3 rectangular regions. what dimension of the corral maximize the enclosed area? what is the area of that corral?
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