Question 161636: CAN SOMEONE HELP ME WITH THIS PROBLEM:
A rectangular enclosure must have an area of at least 1800 yd^2. If 180 yd of fencing is to be used, and the width cannot exceed the length, within what limits must the width of the enclosure lie?
a. 45
b. 30
c. 0
d. 30
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! based on the problem statement, it looks like the area can be anything at or above 1800 square yards.
the perimeter, however, must be 180 yards.
working with the perimeter, it looks like W can't be > 45 yards since that would be the point where it was equal to L and the perimeter was 180 yards.
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equation for perimeter is
2*L + 2*W = 180
if L and W both = 45, then
2*45 + 2*45 = 180
W can be less than that but no more than that.
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is there a lower limit?
looks like the area will determine the lower limit since there would be no lower limit (except 0) if we just looked at the perimeter.
L could be 90 and W would be 0 so any value of L smaller than 90 would allow there to be an enclosure with a perimeter of 180.
the area, however, would not be >= 1800 square yards.
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formula for area of enclosure is A = L * W
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solving for L yields
L*W = 1800
L = 1800/W
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formula for perimeter of enclosure is P = 2*L + 2*W = 180
substituting 180/W for L and equation becomes
2*(180/W) + 2*W = 180
multiplying both sides of the equation by W and it becomes
2*(1800) + 2*W^2 = 180 * W
after algebraic manipulations, this equation becomes
2*W^2 - 180*W + 3600 = 0
solving for W, we get
W = 30
W = 60
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i'm assuming you know how to solve the quadratic so i skipped the details of how i got that answer.
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W can't be > 45, so W = 60 is out.
that leaves W = 30.
for an exact area of 1800, W = 30 would be good.
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if W was less than 30, we could still get an area of at least 1800, but that would not satisfy the perimeter equals exactly 180 requirement.
example:
let W = 29
to satisfy the perimeter requirement, L would have to be (180 - 2*29)/2 which would be 61.
with L = 61 and W = 29, the perimeter is 180.
but,
the area is 61 * 29 = 1769.
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looks like D is the correct answer.
30 <= W <= 45.
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i say <= because i believe <= meets the requirement.
area has to be at least 1800 square yards.
30 * 60 = 1800 satisfied that part of the equation.
perimeter has to be exactly 180 yards.
2*30 + 2*60 satisfies that.
if W = 45 and L = 45, then
45 * 45 = 2025 > 1800 so area is satisfied.
2*45 + 2*45 = 180 so perimeter is satisfied.
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my answer would be
30 <= x <= 45
best answer closes to that would be d.
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one last validity test.
i picked a number for W that was > 30 and < 45 at random.
that number is, let's say, 35.
if i use 35 as W, the perimeter has to be 180, then L has to be 55.
if L is 55, and W is 35, then A = 1925.
looks like we're ok with W >= 30, and W <= 45.
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