a rectangular field is to be enclosed with a fence?

Question

a rectangular field is to be enclosed with a fence. One side of the field is against an existing fence, so that no fence is needed on that side. IF material for the fence cost $2 per foot for the 2 ends and $4 per foot for the side parallel to the existing fence, find the dimensions of the field of largest area that can be enclosed for $1000

Answer 1

Let's measure length on the side parallel to the new fence, and width on the ends.

Your cost:
C = $2 * ends_total + $4 * length
C = 2 * 2W + 4L
C = 4W + 4L

Knowing that the cost is limited to $1,000, you can solve for W:

C = 4W + 4L
1000 = 4W + 4L
250 = W + L
W = 250 - L

The area is just width times length, but you can substitute for W per the above:

A = W*L
A = (250 - L) * L
A = 250L - L^2

The maximum area occurs when dA/dL = 0:

dA/dL = 0
d(250L - L^2)/dL = 0
250 - 2L = 0
2L = 250
L = 125

Once you know the length, you can compute the width:

C = 4L + 4W
1000 = 4L + 4W
1000 = 4(125) + 4W
1000 = 500 + 4W
4W = 500
W = 125

So... the maximum area that you can create is a 125 x 125 square, which would be 15,625 square feet.

The cost would be $2*125 = $250 for each end, and $4*125 = $500 for the width.

Note that this is a lot longer than most of the other responses, however this calculation is the necessary one. One can't just assume that a square (the shape of maximum area for a fixed perimeter) is the right answer, as some others did.

If the costs were different (for example $2 per foot for the width), the correct answer would NOT be a square, because the perimeter would change depending on one's choices for length and width. In this case the twice-as-expensive length compensates for use of the other fence, and you end with $4 total per foot of length or width -- meaning that adding a foot to the length requires removing a foot from the width, leaving the perimeter the same. If width were $2 instead, then adding a foot of length ($4) would require removing TWO feet of width ($2*2), and the problem would not be one of maximizing area under constant perimeter.

Answer 2

When you do optimization problems, the first thing is to ask yourself, what is being optimized? In this problem, you're asked for the largest area. Ok, so that means you need a formula for area. What shape are you working with? In this problem, a rectangular field. So say it has length L and width W, then its area is

A = L*W

Next step is to find the constraint equation so you can solve for L in terms of W (or vice versa). The constraint here is that your cost is limited to $1000. Call cost "C".

C = 1000

But what is a formula for C? Well that's where all the stuff in the middle comes in. Fence material costs $2 per foot for the 2 ends and $4 per foot for the parallel side. So if each end is W feet long and the parallel side is L feet wide, the cost is:

C = 2*W + 2*W + 4*L

(2 dollars per W feet for the two sides and 4 dollars per L feet for the one side)

Now since C = 1000, we get

1000 = 2*W + 2*W + 4*L

which simplifies to

1000 = 4*W + 4*L

now you solve for W in terms of L:

1000 - 4*L = 4*W

250 - L = W

and plug back into original equation for W:

A = L*W
A = L*(250 - L)
A = 250*L - L^2 (simplifying)

Now that you have the equation in terms of one variable only, do your standard optimization stuff. Take 1st derivative of A,

A' = 250 - 2*L,

set it equal to 0 and solve for L to get the critical values. (One in this case.) Do either the first or second derivative test to be sure it's a max, and that'll be the L-dimension for your answer. You're asked to give both dimensions, so once you get the L which maximizes area, go back to

250 - L = W

and plug in the maximal L you found to get the W dimension as well.

Answer 3

the largest dimensions you can have are 125ft x 125ft.

using the equation 1000 = 2x + 2x + 4y or 1000 = 4x + 4y

in that case just start plugging in values.

though a 125 x 125 field = 15,625 square feet it doesn't look like a traditional rectangle. If it needs to be a traditional rectangle then you could go with 124 x 126 = 15,624 square feet.

Answer 4

1000 = 2*(Length + Length) + 4*(Width)

1000= 2*(2L) + 4W = 4L + 4W

250 = L + W

L=125'
W=125'

You get a square/rectangle 125' on each side.

***In response to answer above, a square is just as much of a rectangle as every rectangle ever to exist. There is no such thing as a traditional rectangle. The only answer is 125' x 125'. If you answer 124' x 125' you will be wrong.

Answer 5

let area = L * W [ L & W are variables]
by differentiation , the largest area is when L = w

SO , the largest area that can be enclosed is a square of side = x ft
the total cost is 2x+2x+4x = 1000
8x = 1000
x= 125 ft

Answer 6

Where x is the length of one of the fence segments perpendicular to the existing fence, and y is the length of the fence parallel to the existing fence:

4x+4y=1000
x+y=250
xy=125^2
x=125
y=125
125'x125' fence, encloses 15,625 sq feet

Answer 7

a ($2/ft)
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|``free``````````````| b ($4/ft)
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a ($2/ft)


(2 + 2)a + 4b = 1000
4a + 4b = 1000
a + b = 250
a = 250 - b

Find a and b such that a + b = 250 and area A is max.

A = ab
= (250 - b)b
= 250b - b^2

dA/db = 250 - 2b

When A is max, dA/db = 0,
Thus 250 - 2b = 0
b = 125

a = 250 - 125 = 125

Therefore, dimensions of field of largest area = 125ft x 125ft

Answer 8

500 ft