SOLUTION: A rectangular field will be fenced on all four sides. There will also be a line of fence across the field, parallel to the shorter side. If 900m of fencing are available, what di

Question 1183771: A rectangular field will be fenced on all four sides. There will also be a line of fence across the field, parallel to the shorter side. If 900m of fencing are available, what dimensions of the field will produce the maximum area? What is the maximum area?
Write a function f(x) to find the area, if x is the length of the fence going across the field.
What is the maximum area that can be enclosed by using all of the fencing material and the fence lengths?
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Draw this.
Long side is x and sum is 2x
Short side is y and sum is 3y, since there is an extra fence
We know that each short side fences is (1/3) (900-2x) or 300-(2/3)x
the area is x*(300-(2/3)x)=300x-(2/3)x^2. This is f(x)
and that has to be maximized
the vertex for this negative quadratic is x=-b/2a or -300/(-4/3)=225 m
so the dimensions are 225 m x 150 m, and the area is 33750 m^2.
The length is 225 m and each width is 150 m
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The solution from the other tutor is fine.

If you are in any position where you encounter this same kind of problem often, you can answer them in a tiny amount of time using this fact:

The area is maximized when equal amounts of fencing are used for the two lengths as for the three (or any number) of widths.

Here is a proof....

Let the number of fence portions across the width of the field be A and the total amount of fencing be B. Then

The area is

The maximum value of that quadratic expression is when x=-b/2a:

Then the total amount of fencing to be used for the two lengths is

That says that half of the total amount of fencing B should be used for the two lengths; and that of course means the other half should be used for the widths -- so equal amounts of fencing are used for the lengths and the widths.

So armed with this knowledge, this particular problem can be solved in very little time as follows:

Total fencing 900m
amount to be used for lengths: 900/2=450m
length of field: 450/2=225m
amount to be used for widths: 900/2=450m
width of field: 450/3=150m

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