Question

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.

Answer 1

set up a couple equations.
x+y=30
\[10x+\sqrt{y}\]=f(x,y)

What math is this for?

Answer 2

algebra

Answer 3

first off, I read that wrong.

Answer 4

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

Answer 5

so how do you find the two numbers?

Answer 6

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.

Answer 7

yes I see so far

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

i believe integers

Answer 15

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.

Answer 16

10 and 20?

Answer 17

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

Answer 18

ok thank u

Answer 19

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

Answer 20

where 25 is the one you multiply by 10.

Answer 21

25 and 5? ok thank you!

Answer 22

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.