finding length of fence

Question

A farmer is setting up separate pens for livestock by fencing off a rectangular area and running fence to create four parallel pens. If the farmer has a total of 1000 meters of fence, what is the largest area that can be contained?

Answer 1

This is very common Calculus problem: optimize some quantity, subject to some constraint.

Here we want to optimize (in this case maximize) is the area of the field, subject to the constraint of having only so much fencing.

The solution method is to take the thing you want to optimize and write it as a function of something you can control -- or, at least, control as much as the constraint lets you. Then take the derivative of the function, set it equal to 0, solve to find the value of the variable, and plug that into the function to find the optimum.

In this problem, what we can control is the length of the pieces we cut the fencing into, in order to build the complete fences that split the field into the four parallel pens.

The entire field has a length which we don't know, so let's call the length "x". It has a width which we don't know, so let's call the width "y". And it has an area which we don't know, so let's call the area "A".

But we do know that the area of a rectangle is the length times the width. That is

A = xy (Equation #1)

So, in order to fence the perimeter (outside) of the field, we need two fences each equal to the length, and two fences each equal to the width. That is, so far the amount of fence we need, just for the field itself, is 2x + 2y.

Let's make the four parallel pens with their lengths parallel to the length of the field. If you draw a picture, you'll find that this can be done with three more fences inside the field. The length of each of those fences will also be y, the length of the field.

So the total length of all the fences (call it "L") will be 2x + 2y + 3y, or

L = 2x + 5y (Equation #2)

To get the biggest possible area, we should use all the fencing we have. And we know that the total amount of fencing is

L = 1000 (Equation #3)

Combining Equation #2 and Equation #3 gives

2x + 5y = 1000 (Equation #4)

We need to get A as a function of a single variable, so that we can do the calculus. Currently we have A as a function of two variables. So we need to eliminate one of those variables by expressing it as a function of the other variable.

Equation #4 shows a relationship between x and y, so we can use it to find y as a function of x:

2x + 5y = 1000

5y = 1000 - 2x

y = (1000 - 2x)/5

y = 200 - (2/5)x (Equation #5)

Now we can substitute that expression for y into Equation #1:

A = xy

= x[200 - (2/5)x]

A = 200x - (2/5)x^2 (Equation #6)

Now we can take the derivative of the function A in Equation #6

dA/dx = (d/dx)[200x - (2/5)x^2]

= (d/dx)(200x) - (d/dx)[(2/5)x^2]

= 200 - (2/5)(d/dx)(x^2)

= 200 - (2/5)(2x)

dA/dx = 200 - (4/5)x (Equation #7)

Set that equal to 0 and solve for x:

200 - (4/5)x = 0

-(4/5)x = -200

(4/5)x = 200

x = (5/4)200

x = 250 (Equation #8)

Check that this value of x will give us the maximum by plugging it into the second derivative of A:

dA/dx = 200 - (4/5)x

(d^2)a/dx^2 = (d/dx)[200 - (4/5)x]

= (d/dx)(200) - (d/dx)[(4/5)x]

= 0 - 4/5

(d^2)a/dx^2 = -4/5 (Equation #9)

We don't need to plug in x = 250, since there is no x to plug it into! The second derivative simply IS negative, so the point x = 250 will indeed give us the maximum.

We can substitute that value of x into the formula for y, Equation #5:

y = 200 - (2/5)x

= 200 - (2/5)(250)

= 200 - 100

y = 100 (Equation #10)

Now we can substitute the values of x and y into the equation for A, Equation #1:

A = xy

= (250)(100)

A = 25,000 m^2 (Equation #11)

The largest possible field will have an area of 25,000 sq meters.

Answer 2

2x + 5y = 1000; A = xy; x = 500 - 5/2y; A = y(500 - 5/2y) = 500y - 5/2y²; dA/dy = 500 - 5y = 0; y = 100

A'' = - 5 < 0. So, at y = 100 m we have max area A_max = 25000 m²