Question

a dog breeder wishes to create 5 adjacent pens for his dogs. The breeder has 120ft of fencing to use. What is the largest area he could enclose?

Answer 1

I have a feeling your question is somewhat incomplete.

Answer 2

Well the only thing missing is the diagram.
I copied the question word for word

Answer 3

otherwise the area for each pen can be taken to be equal and 120 sqft can be divided by 5.
But that is too simple.

Answer 4

This is calculus, so that seems like it would be way too simple. Also if we did it your way, I think we would be counting each line twice.

Answer 5

A=6w*l

Answer 6

Right. w is the width and l is the length. The largest area to be enclosed wrt height is a constant A given by A=6w*l
Differenciating with l, A is zero (because it's constant)
this 6w+ 6*l*(dw/dl)=0
That gives me dw/dl=-w/l

That is the first step.

Answer 7

Fencing is done around the perimeter, and the perimeter for 5 pens (now, I trust the owner wants equal areas) is 2[(l/5)+w}=P (I trust the owner wants length equally distributed).

Answer 8

The total perimeter in totality is: the breadth is taken 6 times and length is same, thus
P=120ft(given)=2[6w+l] M'kay?
Now, differenciate both sides wrt l.
That gives you, 6 (dw/dl)+1=0
That is, dw/dl=-1/6

From previous equation, we have dw/dl=-w/l

That is, w/l=1/6
That is, l=6w

Now 2[6w+l]=120
That is, w=5 and l=30
putting the values you get the area.

Answer 9

Wow, Thanks for your help

Answer 10

Just to verify 6(5)*30=900sq ft

Answer 11

Yes.