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Mathematics 24 Online
OpenStudy (hitaro9):

The product of two numbers is 48. Find the sum of the two numbers, S(x), as a function of one of the numbers, x S(x) = Find the positive numbers that minimize the sum and list them in increasing order.

OpenStudy (hitaro9):

I thought it would be x + 48/x But that only leaves 1 positive value for x when I take derivative and set to 0.

OpenStudy (hitaro9):

One of the answers is sqrt 48 Can't figure out the other.

OpenStudy (anonymous):

who said there must be two values where f ' (x) = 0?

OpenStudy (hitaro9):

The online system my teacher is using, webassign There are two blank answers

OpenStudy (hitaro9):

And it says they both have to be positive. -sqrt48 gave an incorrect answer.

OpenStudy (anonymous):

If one number is x, the other number is 48/x. If one number is -x, then the other number is -48/x. So one pair can be both being positive. The second pair is where both numbers are negative.

OpenStudy (hitaro9):

So how do I figure out the positive numbers?

OpenStudy (hitaro9):

I've taken a screenshot of the problem

OpenStudy (anonymous):

You want to minimize the sum...that means you want to find the minimum value for x + (48/x). And you'll need to take the derivative of x + (48/x) and find critical points, etc.

OpenStudy (anonymous):

Don't you get for the derivative 1 - (48/x^2)?

OpenStudy (hitaro9):

Yeah. I only managed to find 2 critical points, 1 of which was negative. I need 2 positive values.

OpenStudy (anonymous):

which is (x^2 - 48)/(x^2)

OpenStudy (hitaro9):

Yeah I did, the only numbers I can think of that can make that true are sqrt48 and -sqrt48.

OpenStudy (anonymous):

so you 3 critical points altogether; x = 0, x = sqrt (48); and x = - sqrt(48).

OpenStudy (hitaro9):

Oh 0? Wouldn't that give an error though?

OpenStudy (anonymous):

You want to find critical points by algebra, not by having to guess/think about it. Set the numerator equal to zero, and denominator equal to zero. Those are the critical points. Now use the First Derivative Test to determine where there's a rel min. Thats what you are looking for.

OpenStudy (anonymous):

0 is still a critical point..because that';s where f ' (x) is undefined.

OpenStudy (anonymous):

A critical point is a point where f ' (x) = 0 or undefined.

OpenStudy (hitaro9):

And that would be considered a minimum? The later parts of the problem are asking what f`(x) would be at that point. How would I answer that?

OpenStudy (anonymous):

No. to find a (rel) min/max, you need to use the First Derivative Test.

OpenStudy (anonymous):

the critical points are the "possible candidates" for a rel min/max.

OpenStudy (hitaro9):

Yeah, I get that, and when I do that I find that 0 is undefined, so that doesn't really work for the problem, right?

OpenStudy (hitaro9):

I'm sorry if I'm asking dumb questions. I'm a bit lost.

OpenStudy (hitaro9):

Again, I did post a screenshot of the question if I'm not making it clear enough.

OpenStudy (anonymous):

so - sqrt(48) is a min

OpenStudy (hitaro9):

Yeah, but its asking for 2 positive mins.

OpenStudy (hitaro9):

When I entered in -sqrt48 it gave me an incorrect answer

OpenStudy (anonymous):

I just read what your screenshot said.

OpenStudy (anonymous):

Why don't you use the First derivative Test to see if - sqrt(48) is a min or not?

OpenStudy (hitaro9):

Well it is, the problem is that it's asking to "Find the positive numbers that minimize the sum "

OpenStudy (anonymous):

seems that - sqrt(48) will not be a min (it will be a max)

OpenStudy (anonymous):

sqrt (48) will be a rel. min. according to the First Derivative Test.

OpenStudy (hitaro9):

Oh. Then yeah. You're right. Still need to find a different minimum

OpenStudy (hitaro9):

Hi bibby. Here's a screenshot of the problem

OpenStudy (hitaro9):

I'm so confused

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