Question 1001543: The question is "Ivan has 80 feet of fence to enclose a rectangular garden. What demensions for the garden give the maximum area?" And I don't know how to solve it so help would be great!
Found 3 solutions by fractalier, Alan3354, ikleyn:
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! The answer is a square, 20 x 20 feet.
If you are in calculus, you would take the derivative and set it equal to zero.
If you are in algebra, you could graph it and look for a maximum.
If you are in neither, you could look at how the area changes as you manipulate the dimensions.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! The question is "Ivan has 80 feet of fence to enclose a rectangular garden. What demensions for the garden give the maximum area?" And I don't know how to solve it so help would be great!
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Max area for a rectangle is a square.
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P = 2L + 2W = 80
L + W = 40
L = 40 - W
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Area = W*L = W*(40-W)
Area = -W^2 + 40W
The max is the vertex, at W = -b/2a
W = -40/-2 = 20
L = 20
Answer by ikleyn(52803)  (Show Source):
You can put this solution on YOUR website! .
Let x = the length of the rectangle and y = its width.
Then x + y = 40 (do you understand why?), and you are asked to find x and y in such a way to maximize the product x*y which is the area.
Express y via x: y = 40 - x, and substitute it into the product:
x*y = x*(40-x)
and find the maximum of the quadratic function f(x) = x*(40-x).
Can you do it yourself?
If not, please let me know, I will help you.
Place your answer into the "Comments from student" section. Do not forget to put the number of the problem (# 1001543) in order for I could identify it.
Good luck!
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