Question

Question #b9993

Answer 1

The total length of the partition should be `125` feet, and the height of each partition should be `50` feet (making each section `50` feet x `31.25` feet). This results in a maximum enclosed area of `6250 " feet"^2`

Explanation:

Without loss of generality let us assume that the pen is divided as shown:

Let us set up the following variables:

`{(x, "Total height of the partition (feet)"), (y, "Total length of the partition (feet)"), (A, "Total Area enclosed by the partition (sq feet)") :}`

Our aim is to find `A(x)`, and to maximize the total area, A, wrt x (equally we could the same with `y` and we would get the same result). ie we want a critical point of `(dA)/dx`.

Now, the total perimeter is given as `500` (constant) and so:

`5x + 2y=500`
`:. 2y=500 - 5x`
`:. y=250 - 5/2x` ..... [1]

And the total Area enclosed by the pen is given by:

`A =xy`

And substitution of the first result [1] gives us:

`A =x(250 - 5/2x)`
`\ \ \ = 250x - 5/2x^2`

We no have the Area, A, as a function of a single variable, so Differentiating wrt `x` we get:

`(dA)/dx = 250 -5x` ..... [2]

At a critical point we have `(dA)/dx=0 =>`

`250-5x = 0`
`:. \ \ \ \ \ 5x = 250`
`:. \ \ \ \ \ \ \ x = 50`

And substituting `x=50` into [1] we get;

`y=250 - 5/2(50)`
`\ \ =250 - 125`
`\ \ =125`

We should check that `x=50` results in a maximum area. Differentiating [2] wrt x we get:

`(d^2A)/dx^2 = -5 < 0` when `x=50`

Confirming that we have a maximum area, given by:

`A = (50)(125) = 6250 " feet"^2`

We can visually verify that this corresponds to a maximum by looking at the graph of `y=A(x)`:
graph{250x - 5/2x^2 [-100, 200, -100, 7000]}