Question

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

Answer 1

You first break down 192 into factors:

Explanation:

`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

There is no maximum sum.

Explanation:

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`