# Maxima and Minima application question!

1. Aug 27, 2010

### violetskies

1. The problem statement, all variables and given/known data
A farmer is planning to build six adjoining rectangular pens of equal size to house his hens, as shown in the diagram. He only has 180m of fencing, however, and he wants to make the pens as large as possible. Find the maximum area he could make each pen.

The diagram is basically six rectangles joined together, three above three, if you can imagine this.

2. Relevant equations
P= 6x + 4y??

3. The attempt at a solution
I got a little bit confused after beginning with:
P= 6x + 4y
Not sure if I'm headed in the right direction. Could someone help me answer this please? :)

2. Aug 27, 2010

### lanedance

what are x & y? length and width of a single pen?

in that case i get
L = 9x + 8y = 180m

now find the area in terms of x & y, use the above contsraint to substitute for y(x). Then minimise teh resulting equation with respect to x

3. Aug 27, 2010

### violetskies

Thank you for helping!
Well, they don't actually give the length. The only number given is 180m. In class, we were taught to form an expression for the perimeter, then state one variable in terms of the other, substitute to just have one variable, then differentiate and set to 0 to find the max.
Then we would have the area, but I'm confused as to what equation I should use for the perimeter.

There's a clearer idea of the diagram given:
_____
|_|_|_|
|_|_|_|

Kind of badly shown but hey.. close enough :D

4. Aug 27, 2010

### violetskies

Ignore all of that. So I did this:

P = 2y + x
180 = 2y + x
x = 180 - 2y

A = xy
A = (180-2y)y
= 180y -2y^2
dA/dy= 180 -4y
4y = 180
y= 45

x= 180 -2(45)
= 180 - 90
= 90

45 x 90 = 4050

4050 / 6 = 675

Apparently the answer is 112.5m, but I don't understand why I have to divide 4050 by 6 and then that answer by 6 again to get 112.5..?

5. Aug 28, 2010

### lanedance

you haven't set up the problem correctly.. you need to be clear in what x & y are so i understand what you are trying to do

lets let :
x = width of single pen
y = hight of single pen

then the length of the total fence is
L = 180m = 9x + 8y

now the area of a sinlge pen is
A1 = xy

as there are 6 pens the total area is
A = 6xy

now try optimising and see how you go

6. Jul 29, 2011

### Somera.R.23

let x be equal to the width of the pen
let y be the length of the pen
and S be the area of a single pen

there are 6 pen to be fenced for 180 m;
6(2x + 2y) = 180 (eq. 1)
simplify;
6[2(x+y)]=180
12(x+y)=180
x + y = 15
then differentiate;
1 + y' = 0
y' = -1 (we will use it later)

the area of a single pen is equal to S so;
S = xy (length times width) (eq. 2)
differentiate the equation;
S' = xy' + y
make S' equal to zero to maximize the area (MAXIMA/MINIMA HAS BEEN APPLIED) and substitute the computed value of y'.

0 = x(-1) + y
0 = -x + y
x = y (eq. 3)

substitute eq 3 in eq 1

6(2x + 2y) = 180
6(2y + 2y) = 180
6(4y) = 180
24 y = 180
y = 180/24
y = 7.5 m
since x=y then,
x = 7.5 m also.

Solve the area of each pen by substituting the value of x and y into equation 2.
S = xy
S = (7.5 m)(7.5 m)
S = 56.25 m^2

7. Jul 30, 2011

### Ray Vickson

Your constraint 6(2x+2y)=180 is wrong. Each internal segment of fencing bounds two pens, but you are assuming that each pen's side needs it's own, separate piece of fencing.

RGV

8. Aug 1, 2011

### lanedance

guys this post is a year old, probably not worth re-opening