Proving the highest product of two numbers with a fixed sum

Question

How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?

Example:

Given sum equals $10$.
Why is then $5\times5$ more than $6\times4, 7\times 3, 8\times2$, etc.?

Answer 1

Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.

Answer 2

Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $x\cdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=\frac k2 \implies y=\frac k2$.

Answer 3

Sometimes a figure is worth 1000 words.

Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 \leq n \leq 20$?

Answer 4

You want two numbers which give a particular sum and you want to maximize their product:

$$m+n = c \tag{1}$$

$$m(n) = ? \tag{2}$$

You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get

$$m(c-m) = -m^2+cm$$

If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -\frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-\frac{c}{2(-1)} = \frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).

Answer 5

$4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$

Answer 6

$$ \left( \;\left( \frac{a+b}{2} \right) - t \; \right) \; \; \left( \; \left( \frac{a+b}{2} \right) + t \; \right) \; = \; \left( \frac{a+b}{2} \right)^2 \; - \; \; t^2 $$