The Farmer and the Goats

The maximum area confined within 4 straight edges of limited total lengths would be in the form of a square. So, if we only need to provide for 3 of the fence, the maximal area should be in the form of half of that square, with 2 parallel fences perpendicular to the wall and another one parallel to the wall with length twice the other two (or the sum of the other two, and equal to half of the provided fencing).

Let the width of the fence be $x$ . Then, since the total length of the fence is $200m$ , the length of the fence (i.e. the third side only) will be $200-2x$ .

The area then can be expressed as

$A = x(200 - 2x)$

$A = 200x - 2x^{2}$

We then differentiate to find:

$\dfrac {dA} {dx} = 200 - 4x$

Stationary points occur when $\dfrac {dA} {dx} = 0$ , therefore

$200 - 4x = 0$

$x = 50$

It follows that the maximum area of the field enclosed is $50 \times 100 = 5000m^{2}$

Since each goat needs $100m^{2}$ of space in order to live healthily, the maximum number of goats the field can sustain is therefore

$\dfrac {5000} {100} = 50$ .