Maximum and minimum points 
Two real numbers $x$ and $y$ are such that $2x + y = 100$. Find the maximum value of the product of the two numbers.

Not sure how to attempt this. It is in differentiation questions but not sure if it actually includes any. Any suggestions?
 A: Hnt: write $$xy$$ as $$x(100-2x)=-2x^2+100x$$ This is a quadratic function in $x$
A: Note that by AM-GM, assuming $x,y>0$
$$\frac{2x+y}2\ge\sqrt{2xy}\implies xy\le1250$$
A: Use A.M. >= G.M. inequality, 
$$\cfrac{2x+y}{2} >= \sqrt{2xy} \implies xy <= 1250$$  
Clearly when $xy$ is maximum $x>0$ and $y>0$.
A: $2x+y=100$
$\implies y=100-2x$
consider the equation $S= xy$  , this is the equation to be maximized
$S= x(100-2x)$
$S= 100x-2x^2$
$\frac{dS}{dx} = 100 - 4x$
set $\frac{dS}{dx}=0, \implies 100 -4x = 0 \implies x = 25$
$\frac{d^2S}{dx^2} = -4 \lt 0 $
second derivative is negative ,so it is maximum at $x=25$
when $x= 25\implies y =50$
Therefore  the 2 numbers are 50 and 25  
A: You want to find the maximum value of $P=x\cdot y$.  So find a function $f$ equal to the product, and solve for the derivative equals $0$.
The problem is that the function $f$ must be in terms of one variable and $P=x\cdot y$ is in terms of two variables.
S we must express $y$ in terms of $x$ (or equivalently vice versa).
Now $2x + y = 100$ so $y = 100 -2x$.  That's it.
So $f(x) = P = x\cdot y = x(100-2x)$
So now we just have to solve $f'(x) = 0$.
....
Of course we are taking it on faith the result will be a maximum.  We should also verify that $f''(x) < 0$ to assure it is not a minimum or a saddle point.
.... Um, BTW, why is your title about maximum AND minimum.  $f'(x)=0$ has only one solution.  There is no minimum product.  If we take $x$ to be as large or as negative as we like $x(100 - 2x)$ will be unbounded negative.  So there is no minimum. 
