Thread: Derivative Quiz Bonus

The product of two positive numbers is 588. The sum of the first and three times the second is minimized. Find the two numbers.

a) 42 and 14
b) 28 and 21
c) 49 and 12
d) 84 and 7
e) Both numbers are 14sqrt(3)

Our teacher gave us a take home quiz on derivatives that I blew through no problem, but I have no idea how to do this bonus. I've never seen the term "minimized" before but I'm guessing it has something to do with a relative minimum.

Originally Posted by vesperka

The product of two positive numbers is 588. The sum of the first and three times the second is minimized. Find the two numbers.

a) 42 and 14
b) 28 and 21
c) 49 and 12
d) 84 and 7
e) Both numbers are 14sqrt(3)

Our teacher gave us a take home quiz on derivatives that I blew through no problem, but I have no idea how to do this bonus. I've never seen the term "minimized" before but I'm guessing it has something to do with a relative minimum.

indeed it does have to do with that. set up your equations:

let the two numbers be and , x is the first number, y is the second

we have that

we are also given a formula to be minimized, let's call it

so,

replace x with the formula from the first equation, we have:



Now, minimize this function, that is, find it's local minimum, the value for y that gives that is one of the numbers you seek. use it to find the other number (Hint: using the quotient rule to differentiate the first term is a waste of time)

Thanks for your help so far, very easy to follow

However, I'm stuck on the last step (it's my first day back from winter vacation heh).

To find a local minimum, normally I take the derivative, find the critical points, then create a sign chart to see a sign change from negative to positive.

Now, I'm not familiar with derivatives involving y. I think it has something to do with implicit differentiation which we have yet to actually implement. Ignoring this though, I would find the derivative to be:

S' = -558y^-2 + 3

Instead of using the quotient rule, I rewrote 558/y as 558y^-1. If I'm correct up to here (which I'm guessing I'm not because I'm not sure how to find the derivative with respect to y) what do I do next?

Originally Posted by vesperka

Thanks for your help so far, very easy to follow

However, I'm stuck on the last step (it's my first day back from winter vacation heh).

To find a local minimum, normally I take the derivative, find the critical points, then create a sign chart to see a sign change from negative to positive.

Now, I'm not familiar with derivatives involving y. I think it has something to do with implicit differentiation which we have yet to actually implement. Ignoring this though, I would find the derivative to be:

S' = -558y^-2 + 3

Instead of using the quotient rule, I rewrote 558/y as 558y^-1. If I'm correct up to here (which I'm guessing I'm not because I'm not sure how to find the derivative with respect to y) what do I do next?

implicit differentiation not needed here. we are working in one variable, namely y, and we are differentiating with respect to y, so it's fine. yes, you did the differentiation correctly. now continue. find the critical point(s) of S by setting S' = 0 and solving for y (usually sign charts are not necessary with these kinds of questions, they are called "Optimization problems" by the way).

Oh haha, I'm such a chump. I tried using my graphing calculator and had it setup all wrong, but when I did it out by hand it was much easier. I solved for y in that equation and got 14, which means the two numbers are 42 and 14.

Thanks for the help, extra credit here I come!

Originally Posted by vesperka

Oh haha, I'm such a chump. I tried using my graphing calculator and had it setup all wrong, but when I did it out by hand it was much easier. I solved for y in that equation and got 14, which means the two numbers are 42 and 14.

Thanks for the help, extra credit here I come!

no problem, you're welcome

good luck with your class

Originally Posted by vesperka

Oh haha, I'm such a chump. I tried using my graphing calculator and had it setup all wrong, but when I did it out by hand it was much easier.

This happens much more often than students would like to believe.

-Dan