## anonymous 5 years ago Can someone help with this?: Find two numbers whose sum is 30, such that the sum of the square of one number plus ten times the other number is a minimum.

1. anonymous

set up a couple equations. x+y=30 $10x+\sqrt{y}$=f(x,y) What math is this for?

2. anonymous

algebra

3. anonymous

first off, I read that wrong.

4. anonymous

sorry about that. Anyway, the equation should really be 10x+y^2->minimum

5. anonymous

so how do you find the two numbers?

6. anonymous

Sorry, I forget how to do it with what you know in algebra, but let's see if I can figure it out as I go along. Using those equations, you also know that 30-y=x and can substitute that in.

7. anonymous

yes I see so far

8. anonymous

so 10(30-y)+y^2 needs to be minimized.

9. anonymous

which means that 300-10y+y^2 needs to be minimized.

10. anonymous

ok so the next step would be setting it to zero?

11. anonymous

I guess. Sorry, I don't remember how to do this the algebra way.

12. anonymous

well I don't really get anywhere doing that anyway. How would you do it?

13. anonymous

do the numbers have to be integers? I would use lagrange multipliers, which I'm pretty sure you haven't learned about, and you don't need for the problem.

14. anonymous

i believe integers

15. anonymous

well to minimize them, 10x would have to be as close to y^2 as possible, so you can just guess and check if they have to be integers to see what gets you the closest.

16. anonymous

10 and 20?

17. anonymous

I don't know. Try asking the question again to see if anyone can do it better.

18. anonymous

ok thank u

19. anonymous

actually, using my way, which I'm not going to explain because it's not the way you're supposed to do it, I got 25 and 5

20. anonymous

where 25 is the one you multiply by 10.

21. anonymous

25 and 5? ok thank you!

22. anonymous

yeah, and you can check that by seeing that no other numbers get you a higher number when you multiply one by 10 and square the other, but I'm almost positive that the answer is 25 and 5.

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