- #1
moo5003
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Question:
A total of x feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?
Answer:
My first guess is x^2 / 9 if you assume the yard is square. I'm a little unsure of my answer since for either the horizontal or vertical (depending on how you set up your yard) you only have to fence one side of the yard, thus it may be beneficial to extend that side since you are required to fence less of it and increase overall area. Can anyone please explain their answer in comparison to mine?
What I did:
A= Area
X= Fencing in feet
L= Length
W= Width
A= L*W
X = 2W + L
Thus:
X = 2W + A/W
XW - 2W^2 = A
Which takes a maximum at W = X/4
Thus area should be maximized at L = X/2, W= X/4 thus maximum area is X^2/8
Questions:
This is for the subject Math GRE (Question 13 on the prep exam), how would any of you do this quickly? My way seemed to take more time then I would want to spend on something like this.
A total of x feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?
Answer:
My first guess is x^2 / 9 if you assume the yard is square. I'm a little unsure of my answer since for either the horizontal or vertical (depending on how you set up your yard) you only have to fence one side of the yard, thus it may be beneficial to extend that side since you are required to fence less of it and increase overall area. Can anyone please explain their answer in comparison to mine?
What I did:
A= Area
X= Fencing in feet
L= Length
W= Width
A= L*W
X = 2W + L
Thus:
X = 2W + A/W
XW - 2W^2 = A
Which takes a maximum at W = X/4
Thus area should be maximized at L = X/2, W= X/4 thus maximum area is X^2/8
Questions:
This is for the subject Math GRE (Question 13 on the prep exam), how would any of you do this quickly? My way seemed to take more time then I would want to spend on something like this.