Math Is Fun Forum

sirsosay

Member

Registered: 2005-12-15

Posts: 11

Another problem.. :)

I don't know if there is a limit to the amount of questions you can ask here, but so far the help has been really great.

The sum of the squares of two positive numbers is 200.  The minimum product of these two numbers is?

So I set up an equaiton...

a²+b²=200

So do I need to substitute one variable in for the other and solve the derivative of ab?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Another problem.. :)

Yea, so expand on that fact:

a^2 + b^2 = 200

So a*b = √(200 - b^2) * b

So you want to find the minimum of that function.  Take a derivative, find the critical points, and test each one.

Hint: You should come up with 2 critical points, only one will be within your domain (i.e. positive).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Another problem.. :)

I drew a rectangle with a diagonal in it and tried to keep the
diagonal a constant # like sqrt(200), but any # is fine.
Then I realized if height or width approaches 0, but is just
tiny bit over zero like 0.01, then you will get the
minimum area or product, but still have a diagonal length
of sqrt(200).   So the minimum product approaches zero,
I think.

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igloo myrtilles fourmis

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Another problem.. :)

Ah, I forgot to do the second derivative test and found a maximum instead of a minimum.  Then the only critical point you are left with is 0, but 0 isn't positive.  So in this situation, you want to find a number closest to this minimum that is in your domain.  In this case, it would be:

lim as x->0+ of x, which goes towards 0.  Are you sure you didn't mean maximum?  That's how this question is usually given.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."