# How do you find two positive numbers whose product is 192 and the sum is a maximum?

Jul 3, 2015

You first break down 192 into factors:

#### Explanation:

$1 \cdot 192$ sum=193
$2 \cdot 96$ sum=98
$3 \cdot 64$ sum=67
etc.
You will see that $1 \mathmr{and} 192$ gives the highest sum.

This only works if by "number" you mean a positive integer.

(in Dutch we have two words for 'number':
"nummer" is allways a positive integer
"getal" may be anything, like $3.345 \cdot {10}^{-} 3$)

Jul 4, 2015

There is no maximum sum.

#### Explanation:

If we initially call the numbers $x$ and $y$, then we have

$x y = 192$, so $y = \frac{192}{x}$

The sum is then $f \left(x\right) = x + y = x + \frac{192}{x}$ with domain $\left(0 , \infty\right)$.

Maximize $f \left(x\right) = x + \frac{192}{x}$ on the domain $\left(0 , \infty\right)$.

$f ' \left(x\right) = 1 - \frac{192}{x} ^ 2 = \frac{{x}^{2} - 192}{x} ^ 2$

$0$ is not in the daomain, so the only critical numbers are the zeros of $f '$.

$f ' \left(x\right) = \frac{{x}^{2} - 192}{x} ^ 2 = 0$ at

${x}^{2} - 192 = 0$, so $x = \pm \sqrt{192} = \pm 4 \sqrt{12}$.

The restricted domain eliminates the negatve value. The only critical point of interest is $\sqrt{192} = 4 \sqrt{12}$.

Do not assume that this is the answer!

$f ' ' \left(x\right) = \frac{384}{x} ^ 3$ and $f ' \left(\sqrt{192}\right)$ is positive, so $f \left(\sqrt{192}\right)$ is a minimum. We want a maximum.

Now the bad news. There is no maximum sum.
We could take
$1 \times 192$ whose sum is $193$

$\frac{1}{2} \times 384$ whose sum is $384 + \frac{1}{2}$

$\frac{1}{4} \times 768$ whose sum is $768 + \frac{1}{4}$

and so on.

In fact ${\lim}_{x \rightarrow 0} \left(x + \frac{192}{x}\right) = \infty$