A farmer has 1000 meters barbed wire with which he is to fence off three sides of a rectangular field, the fourth side being bounded by a straight canal. how can the farmer enclose the largest field?

Answer 1

make a sketch,
let teach of the three sides be x ft
let the length of the side parallel to the canal be y

y + 3x = 1000
y = 1000 - 3x

area = A = 3xy
= 3x(1000 - 3x)
= -9x^2 + 3000x

If you know calculus ...
dA/dx = -18x + 3000 = 0 for a max of A
18x = 3000
x = 3000/18 = 500/3
then y = 1000 - 3(500/3) = 500
max area = 3xy = 3(500/3)(500 = 250000

The length is 500 m and each of the 3 shorter sides is 500/3 m

if no calculus....
complete the square on
A = -9x^2 + 3000x
= -9(x^2 - (1000/3)x +250000/9 - 250000/9)
= -9(x - 500/3)^2 + 250000
So the max of A is obtained when x = 500/3 (as above)
for a max area of 250000

or

for A = -9x^2 + 3000x
the x of the vertex is -b/(2a) = -3000/(2(-9)) = 500/3

...continue as above

Answer 2

I have a different interpretation of the problem.

If I were to assume "three sides" as one length (parallel to the canal) and the two other sides perpendicular to the canal, then we are only fencing three sides on land, and the fourth side is bounded by the canal.

Let x=width (side perpendicular to canal)
Area, A(x)
=x*(1000-2x)
=1000x-2x²

No calculus, complete square:
A(x)=-2(x-250)²+125000
meaning the maximum is at x=250.

With calculus,
A'(x)=1000-4x=0 => x=250

So the two short sides are 250 m, and the long side is 500 m. The fourth side is assumed to be bounded by the canal.

Answer 3

After reading the question again more carefully, I have to agree with MathMate's interpretation.

There is quite a variety of these type of questions and without a diagram I just jumped to the interpretation I used.