SOLUTION: please help me What is the largest area that can be enclosed by fence? We have 180m of fence and there is a preexisting fence on one side. Round to 3 significant digits.

Question 1163079: please help me
What is the largest area that can be enclosed by fence? We have 180m of fence and there is a preexisting fence on one side.
Round to 3 significant digits.

Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.

Since one side is just fenced, the total length of the three other sides of the rectangle, W, L an W is L + 2W = 180 meters. Hence, L = 180 - 2W meters. Area = Length * Width. Substitute (180-2W) for L: A = W(180 - 2W) (1) A = -2W^2 + 180W. It is a quadratic function. It has the maximum at x = -b/(2a), where "a" is the coefficient at the quadratic term and "b" is the coefficient at the linear term, according to the general theory. (See the lessons - HOW TO complete the square to find the minimum/maximum of a quadratic function - Briefly on finding the minimum/maximum of a quadratic function in this site). In your case, the maximum is at W = = = 45. So, W = 45 meters is the width of the rectangle for the max area. Then the length is L = 180 - 2W = 180 - 2*45 = 90 meters. So, the dimensions of the rectangle are 90 m and 45 m. The max area is A = L*W = 90*45 = 4050 square meters. ANSWER.

Solved.

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My other lessons in this site on finding the maximum/minimum of a quadratic function are
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola

    - A rectangle with a given perimeter which has the maximal area is a square
    - A farmer planning to fence a rectangular garden to enclose the maximal area (*)
    - A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
    - Finding the maximum area of the window of a special form
    - Using quadratic functions to solve problems on maximizing revenue/profit

    - OVERVIEW of lessons on finding the maximum/minimum of a quadratic function

The most relevant to your problem is the lesson marked (*) in the list.

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My notice after reading the post by @Theo.

            The original formulation in the post is absolutely and crystally clear.

            The interpretations that @Theo tries to make, to discuss and to use, all are irrelevant and do not fit to the problem.

            Simply ignore his post, for your safety.

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
you have a total of 180 meters of fence and there is a pre-existing fence on one side.

if i assume that the 180 meters of fence is the length of the existing fence that will be taken down and used to enclose the area, then this means your perimeter will be equal to 180 meters.
the largest area is when the area is a square, so you need to do the following.
s = the length of a side of the enclosed area.
the perimeter is equal to 4 * s
perimeter = 4 * s
since perimeter = 180, then you get 180 = 4 * s
solve for s to get s = 45
that would be the length of a side of the enclosed area.
the area would be s^2 = 2025 square meters.

if i assume that the 180 meters of fence is new fence that will be attached to the old fence to make the area, then i have to assume there is sufficient length of the old fence to make one side of the enclosure, whatever that length needs to be.
the 180 of new fence will surround 3 sides of the enclosure.
that means 60 meters on a side (180 / 3 = 60)
the fourth side will be 60 meters of the old fence.
the area would then be 60^2 = 3600 square meters.

bottom line:
if the 180 meters is the length of the existing fence, then your maximum area will be 45^2 = 2025 square meters.
if the 180 meters is the length of new fence that will be attached to old fence, then your maximum area will be 60^2 = 3600 square meters.

to convince yourself that a square gives you the maximum area, do the following:
assume total of length and width = 10
possible integer values of length and width and area = length * width would be:
1 * 9 = 9
2 * 8 = 16
3 * 7 = 21
4 * 6 = 24
5 * 5 = 25 ***** maximum area when length = width
6 * 4 = 24
7 * 3 = 21
8 * 2 = 16
9 * 1 = 9

if you were told the length of the existing fence, then the problem would need to be calculated differently.

i think the maximum enclosed area would still be a square, but let's see if this makes sense.
if the existing fence is at least 60 meters, then the above second assumption applies.

assume the existing fence is 30 meters.
now you have a total perimeter of 30 + 180 = 210 meters.
divide that by 4 to get s = 52.5 meters.
the largest area would then be 52.5^2 = 2756.25 square meters.
your old fence length would be 30 meters.
your new fence length would be 3 * 52.5 meters plus (52.5 - 30) = 22.5 meters.
3 * 52.5 + 22.5 = 180 meters of new fence connected to 30 meters of old fence.

without knowing the length of the old fence, the first two assumptions above would be the only logical assumptions i think i could make.

take your pick as to which of those assumptions you think the person who provided you with the probleman meant.

let me know if a difference assumption needed to be made.

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