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?
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
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.
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.
To enclose the largest field using 1000 meters of barbed wire with three sides, the farmer needs to determine the dimensions of the rectangle that would result in the largest area.
Let's break down the problem:
1. Assume the length of the field parallel to the canal is 'l' meters.
2. The width of the field perpendicular to the canal can be denoted as 'w' meters.
3. The two ends (length) and one side (width) of the rectangle need to be enclosed by the barbed wire, while the side bounded by the canal does not.
To maximize the area, we need to find the values of 'l' and 'w' that satisfy the condition of using all the barbed wire (1000 meters).
Since the barbed wire is used for three sides (2 lengths + 1 width), the total length of the barbed wire used can be calculated as follows:
2l + w = 1000
Since we want to maximize the area, which is given by A = l * w, we need to express one variable in terms of the other and substitute it into the area equation.
From the previous equation, we can solve for w:
w = 1000 - 2l
Substituting this value of 'w' into the area equation:
A = l * (1000 - 2l)
Now, we have an equation for the area 'A' solely in terms of 'l'.
To find the maximum area, we need to differentiate the area equation with respect to 'l', set it to zero, and solve for 'l'.
Differentiating:
dA/dl = 1000 - 4l
Setting it to zero:
1000 - 4l = 0
Solving for 'l':
4l = 1000
l = 250
Substituting this value of 'l' back into our equation for 'w':
w = 1000 - 2(250)
w = 500
Therefore, to enclose the largest field using 1000 meters of barbed wire, the farmer should create a rectangle with dimensions of 250 meters (length) and 500 meters (width).