please help, area problem?

Question

you have 100 meters of fencing material to enclose a rectangular plot.

a) determine the dimensions which maximize the enclosed area.

b) if you double the amount of fencing, what affect does this have on the maximum possible area enclosed? explain.


i have NO earthly idea what this means or is asking. please help me. thank you very much!

Answer 1

100 = 2(L + W)
or 50 = L + W --- (1)

A = L * W
A = L * (50 - L)

Differentiate both side w.r.t L
dA/dL = 50 -2L
To find maximum dA/dL = 0
or 50 -2L = 0
or L = 25

if fencing is 200
200 = 2(L + W)
100 = L + W
A = L * W
A = L * (100 - L)
dA/dL = 100 - 2L = 0
or L = 50

Answer 2

a) We want to maximize the area of a rectangle.

Let x and y be the sides of a rectangle

Area = A = xy

we know the perimeter = 2x+2y =100 or x+y =50
or y =50-x

A = x(50-x) ;

let's differentiate this to get
A' = (50-x) -x =50-2x = 2(25-x)

find the roots of A'
x =25

hence y =25
maximum enclosed area = 25^2 =625 m^2

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b) if you double the amount of fencing, what affect does this have on the maximum possible area enclosed? explain.

double perimeter =200m

x+y =100 ~~~> y =x-100

A = x(x-100)

A' = -100+2x =0 if x =50

hence y =50
A= 2500m^2

Conclusion: doubling the perimeter of fence gives us 4 times the enclosed area

which is expected since a perimeter is of dimension 1, whereas the area is of dimension 2

Good luck

Answer 3

Hello

let w = width
let l = length

so the area is l*W

The perimeter is 2l + 2w = 100

so solving for w we have w = (100 - 2l) /2 = 50 - l

Putting this into the first equation we have.

l*(50-l) = area
50l - l^2 = area

Now take the derivative of this we get
da/dl = 50 -2L setting this equal to 0 gives us 0 = 50 -2l so l = 25. This means that w = 25 and we have a square.
25 meters on a side.

Second part:
Double the perimeter and we have 2l + 2w = 200

so solving for w gives us w = 100- l
Area = (100-l)*l = 100l - l^2 so da/dl = 100 - 2l = 0 and then l = 50 instead of 25 since we are doubling the length of a side and the area is side times side we have 2*2 or 4 times the area.

Hope This Helps!

Answer 4

If you have 100m of fence, the perimeter of your rectangle can be no greater than 100m. There are lots of rectangles that would give P = 100, but which has the greatest area? Play with the numbers a bit and you'll see.

P = 2L + 2W. Since P = 100,
100 = 2L + 2W. Dividing by 2,
50 = L + W. Any two numbers that add to 50 would make up the dimensions of your plot. Which has the greatest area?
1m + 49m = 50m. (1m)(49m) = 49m².
2m + 48m = 50m. (2m)(48m) = 96m².
3m + 47m = 50m. (3m)(47m) = 141m².
...
24m + 26m = 50m. (24m)(26m) = 624m².
25m + 25m = 50m. (25m)(25m) = 625m².
26m + 24m = 50m. (26m)(24m) = 624m².

Because of commutability, the areas as the lengths grow larger than 25 meters will be a mirror-image to the ones less than 25m. The answer to part (a) is 25m × 25m.

For the second part, what does doubling the amount of fence do? Instead of 50 = L + W, now you'll work with
100 = L + W. I'll let you figure the dimensions and maximum area. :)

Hope this helps, and best of luck!