Afarmer decides to enclosea rectangular garden, using the side of a barn as one side of the rectangle,what is?

gaiagoddess13: Afarmer decides to enclosea rectangular garden, using the side of a barn as one side of the rectangle,what is?
farmer decides to enclosea rectangular garden, using the side of a barn as one side of the rectangle,what is?
a farmer decides to enclosea rectangular garden, using the side of a barn as one side of the rectangle,what is the maximum area the farmer can enclose with 100ft of fence? what should the dimensions of the garden be to give thisarea?

the maximum area the farmer can enclose with 100ft of fence is?

the dimensionsof the garden to give this are is 50 ft by ?

Answers and Views:

Answer by MathMan TG
If all sides were fenced, maximum area = a square,
with both dimensions contributing equally to the area.

Here we have 2 widths and 1 length,
but we still want both to contribute equally to the area.

So make the sum of the 2 widths (perpendicular to the barn) =
the other side or
2w + n = 100
2w = n
4w = 100
w = 25
n = 50

The dimensions are 50 x 25 = 1250 sq ft.

Try nearby dimensions and see they are smaller:
48 x 26 = 1248
52 x 24 = 1248 also
and the more they differ, the smaller the area
60 x 20 = 1200
70 x 15 = 1050
80 x 10 = 800
and so on
if you make the short side parallel to the barn it’s even worse
30 x 30 = 900
20 x 40 = 800
10 x 45 = 450

Note: probably the way want you to do the problem is:
2w + n = 100
w * n = area to be maximized
n = 100 – 2w
w * (100 – 2w) = 100 w – 2w^2
derivative = 100 – 4w
and if you set that to 0, you get w = 25.

But it’s quicker to remember:
make the dimensions contribute equally to the area,
so make the dimensions equal in total to each other.

This applies as well if there are portions dividing it up into several pieces, too.

In some cases, different weights are given to different segments,
such as cost. Doesn’t matter, include that in the same calculation.