Question
A farmer has 400 feet of fencing with which to build a rectangular corral having two internal dividers both parallel to two of the sides of the corral. What is the maximum total area of such a corral?
I know how to maximize area, but this problem is confusing me because of the internal dividers. A step-by step explanation would help me a lot. Thanks!!
I know how to maximize area, but this problem is confusing me because of the internal dividers. A step-by step explanation would help me a lot. Thanks!!
Answers
when it's all done, you should have the total fence divided equally among the lengths and widths.
So, with the two dividers, you have two long sides of length x, and 4 short sides of length y.
2x+4y = 400
x = 200-2y
a = xy = (200-2y)y = 200y - 2y^2
da/dy = 200-4y
da/dy=0 when y=50
so, x=100
So, 4 short lengths of 50 and 4 long sides of 100
a = xy = 100*50 = 5000
So, with the two dividers, you have two long sides of length x, and 4 short sides of length y.
2x+4y = 400
x = 200-2y
a = xy = (200-2y)y = 200y - 2y^2
da/dy = 200-4y
da/dy=0 when y=50
so, x=100
So, 4 short lengths of 50 and 4 long sides of 100
a = xy = 100*50 = 5000
Thanks!!!
Your answer makes no sense as it confuses the variables and skips steps.
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