Maximum area of a rectangle whose perimeter is $100$. (Derivative Approach) Question:

Find the maximum area of a rectangle whose perimeter is $100$?

My Approach:
(I am using derivative method)
Some Basic rules:
1.) If $f''(x) > 0$ then function has minimum value at $x$
2.) If $f''(x) < 0$ then function has maximum value at $x$
Let the length of rectangle $= x$
and the width of rectangle $= y$
$2(x+y)=100$
$y = 50 - x$
As Area $= x × y$
Put the value of y in above equation
$f(x) = Area = x(50 - x)$
$f(x) = 50x - x^2$
Taking first derivative: $f'(x) = 50 - 2x$
Assuming the first derivative to be equal to zero we get the value of x which is $x = 25$.
Now taking double derivative of above equation and putting the value of $x$ we get $f''(x) = -2$ which is less then zero so the function of Area is maximum hence the Area of rectangle will also be maximum.
Now what approach should i use to calculate Area?
Conclusion: 
Is there a more neat way doing the above question more quickly?
What if the question says to calculate minimum Area?
Help would be appreciated.
Thanks,
(Sorry for bad english and if someone finds any mistake feel free to edit and correct them :) )
 A: We have
$$P = 2(l+w) \implies 100 = 2(l+w) \iff \color{blue}{50 = l+w} \tag{1}$$
$$\color{green}{A = lw} \tag{2}$$
Rewriting $(1)$ in terms of $w$ (you could do so in terms of $l$), you’d get
$$w = 50-l$$
Plugging $w = 50-l$ in $(2)$, you have
$$A = l(50-l)$$
$$\color{purple}{A = 50l-l^2} \tag{3}$$
You’ve gotten it correctly until here. Treat $A$ as a function. The resulting quadratic is concave down, so the vertex is a maximum.
$$l_{vertex} = \frac{-b}{2a} = \frac{-50}{2(-1)} = 25$$
Plugging $l = 25$ in $(1)$, it becomes clear that $w = 25$. Hence, the maximum area occurs when there is a square, so 
$$\boxed{A = s^2 = 25^2 = 625}$$
Using only quadratics is faster than using optimization, but that is of course correct as well:
$$\frac{dA}{dl} = 50-2l$$
$$\frac{dA}{dl} = 0 \implies 50-2l = 0 \implies l = 25$$
from which the same answer is obtained.
In response to your second question, you can’t find a minimum area, because it simply doesn’t exist. (Unless if you set $w = 0$ in which case the area becomes $0$.) A concave down parabola has a maximum, not a minimum. A minimum point is found when the parabola is concave up, which isn’t the case here.
A: Let $a$ and $b$ be  sides-lengths of adjacent sides of the rectangle.
Thus, by AM-GM
$$ab\leq\left(\frac{a+b}{2}\right)^2=\left(\frac{50}{2}\right)^2=625.$$
The equality occurs for $a=b$, which says that we got a maximal value.
A: The area is, as you say, $A(x)=x(50-x)$, a quadratic with roots $x=0$ and $x=50$. The graphic is a parabola simmetric respect to the line $x=\frac{0+50}{2}=25$. Also, the parabola is concave down because the principal coefficent is negative. Then, the maximum is when $x=25$ and the value is $25(50-25)=25^2$.
A: You don't need calculus here, though it does work.
You can do this neatly by noting that $$(l+w)^2-(l-w)^2=4lw$$
You want to maximise $lw$ and $l+w$ is fixed at $50$ so you need $l-w$ as small as possible.
To minimise the area you clearly want $l+w$ and $l-w$ to be as close to one another as possible.
