# The sum of two variable positive numbers is $200$. Find the maximum value of their product.

The sum of two variable positive numbers is $200$. Let $x$ be one of the numbers and let the product of these two numbers be $y$. Find the maximum value of $y$.

NB: I'm currently on the stationary points of the calculus section of a text book. I can work this out in my head as $100 \times 100$ would give the maximum value of $y$. But I need help making this into an equation and differentiating it. Thanks!

-
The answer must be $y=(100)^2$, obtained when the two summands $a, b$ are equal. The maximum must be attained as a special configuration, that is, at a configuration with a particular symmetry. The only such configuration here is $a=b$. Of course this is not a proof; for a formal proof you can either use the GM-AM inequality (like lab bhattacharjee) or techniques from calculus (like Samrat Mukhopadhyay). –  Giuseppe Negro May 31 '14 at 15:00

Let the first number be $x$ and the second number be $z$. We have $$x+z=200\quad\Rightarrow\quad z=200-x.$$ We want to maximize $$y=xz=x(200-x)=200x-x^2.$$ Setting the first derivative equals $0$ yields \begin{align} \frac{d}{dx}\left(200x-x^2\right)&=0\\ 200-2x&=0\\ 2x&=200\\ x&=100. \end{align} Check the second derivative $$\frac{d^2}{dx^2}\left(200x-x^2\right)=-2.$$ Since the second derivative of $y<0$, then $y$ will be maximum at $x=100$. Thus, the value of their product is $y=200x-x^2=\color{blue}{10,000}$.

-
how did you get $y= 200y - y^2$? –  Andros May 31 '14 at 15:05

For positive real $a,b$

$$\frac{a+b}2\ge \sqrt{ab}\implies ab\le\frac{(a+b)^2}4$$

which can also be shown as follows $$(a+b)^2-4ab=(a-b)^2\ge0$$

Alternatively if $\displaystyle a+b=k, ab=a(k-a)=\frac{k^2-(k-2a)^2}4\le\frac{k^2}4$

-

The problem is the following: $$\mbox{Find}\\ y=\max_{x\ge 0}\left(x(200-x)\right)$$ So differentiate the function $f(x)=x(200-x)$ to get $x=100$ as the one that maximizes it (since $f''(100)=-2$. So, $y=100^2$.

-

You call one of your numbers $x$. Write $x$ in the form $100-k$, so the other number is $100+k$ (it does not matter here whether $k$ is zero, or positive, or negative). We have $$z=(100-k)(100+k)=100^2-k^2\le 100^2,$$ with equality happening if and only if $k=0$.

-