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

Question

12017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-03T22:29:41

You first break down 192 into factors:

$1*192$ sum=193
$2*96$ sum=98
$3*64$ sum=67
etc.
You will see that $1and192$ 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 $3.345*10^-3$ )

Answer 2 · 2015-07-04T18:31:06

There is no maximum sum.

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

$xy = 192$ , so $y = 192/x$

The sum is then $f(x) = x+y = x + 192/x$ with domain $(0, oo)$ .

Maximize $f(x) = x + 192/x$ on the domain $(0, oo)$ .

$f'(x) = 1 - 192/x^2 = (x^2-192)/x^2$

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

$f'(x) = (x^2-192)/x^2 = 0$ at

$x^2 - 192 = 0$ , so $x = +- sqrt 192 = +- 4 sqrt 12$ .

The restricted domain eliminates the negatve value. The only critical point of interest is $sqrt192 = 4sqrt12$ .

Do not assume that this is the answer!

$f''(x) = 384/x^3$ and $f'(sqrt192)$ is positive, so $f(sqrt 192)$ is a minimum. We want a maximum.

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

$1/2 xx 384$ whose sum is $384+1/2$

$1/4 xx 768$ whose sum is $768+1/4$

and so on.

In fact $lim_(xrarr0)(x+192/x) = oo$