The Extreme Value Theorem guarantees that a continuous function on a finite closed interval has both a maximum and a minimum, and that the maximum and the minimum are each at either a critical point, or at one of the endpoints of the interval.
When trying to find the maximum or minimum of a continuous function on a finite closed interval, you take the derivative and set it to zero to find the stationary points. These are one kind of critical points. The other kind of critical points are the points in the domain at which the derivative is not defined.
The usual method to find the extreme you are looking for (whether it is a maximum or a minimum) is to determine whether you have a continuous function on a finite closed interval; if this is the case, then you take the derivative. Then you determine the points in the domain where the derivative is not defined; then the stationary points (points where the derivative is $0$). And then you plug in all of these points, and the endpoints of the interval, into the function, and you look at the values. The largest value you get is the maximum, the smallest value you get is the minimum.
The procedure is the same whether you are looking for the maximum or for the minimum. But if you are not regularly checking the endpoints, you will not always get the right answer, because the maximum (or the minimum) could be at one of the endpoints.
(In the case you are looking for, evaluating at the endpoints gives an area of $0$, so that's the minimum).
(If the domain is not finite and closed, things get more complicated. Often, considering the limit as you approach the endpoints (or the variable goes to $\infty$ or $-\infty$, whichever is appropriate) gives you information which, when combined with information about local extremes of the function (found also by using critical points and the first or second derivative tests), will let you determine whether you have local extremes or not. )
Let the available length be $L=100$.
The area of a rectangle is a product of the sides.
Let the one side be $x$m.
After using two such (opposite) sides for the fence,
we have $L-2x$ meters piece left,
from which we need to build two other equal sides,
that is the other side is $\tfrac{L}2-x$.
The area then would be $f(x)=x(\tfrac{L}2-x)=\tfrac{L}2 x-x^2$.
To find such $x$ which maximizes the area,
we need to solve
$f'(x)=\tfrac{L}2-2x=0$, which gives $x=\tfrac{L}4$.
That is, the square with the side $x=25$.
In the second case steps are similar,
except the second size is defined as just $L-2x$.
The function of the area in second case
will differ from the first one only by
the multiplication constant,
so the maximum solution still would be
for $x=\tfrac{L}4=25$
except that in this case it would be a smaller side,
the other side would be $L-2x=50$m.
Best Answer
Use the formula for the area $A$ of a triangle given the length of two sides $a$ and $b$ and the angle between them $C$:
$$ A = \frac 1 2ab\sin C $$
In your case, $a=b=1~\mathrm{km}$.
Since $\frac 1 2ab$ is a constant, in order to maximize $A$ we must maximize $\sin C$. We know from trigonometry that the value of $\sin C$ is at most 1, and $\sin C=1$ when $C=\frac\pi 2=90\deg$.
Since the triangle is also isosceles, that means that we have a 45-45-90 triangle. The hypotenuse is simply $\sqrt 2~\mathrm{km}$ and the height is half that, or $\frac 1{\sqrt 2}~\mathrm{km}$.