Question

optimization:
The sum of two nonnegative numbers is 36. Find the numbers if

a. the difference of their square roots is to be as large as possible.

b. the sum of their square roots is to be as large as possible.

Answer 1

first i know to find the relationsihps so
x + y = 36
so x = 36-y

Answer 2

part b seems like it's the easiest to do, so i started with that one. so to find the optimal i take the derivative of root of x + root of y and set it equal to 0

Answer 3

So far so good.

Answer 4

Right, you have \(x\) and \(36-x\) as your numbers. You also know \(x\geq 0\).

Answer 5

alright so after substiuting in 36 - y for x in the equation i got:
\[\sqrt{36 - y} + \sqrt{y}\]

Answer 6

For part a, we have: \[
f(x) = \sqrt{x}-\sqrt{36-x}
\]And you are trying to find the max.

Answer 7

oh okay, we'll start with part a.

Answer 8

Can you find the critical numbers?

Answer 9

let's see i take the derivative set it equal to 0 and solve for y right?

Answer 10

yes

Answer 11

so for the derivative i got
\[\frac{ 1 }{ 2\sqrt{x} } + \frac{ 1 }{ 2\sqrt{36-x} } = 0\]

Answer 12

You want to multiply both sides by the LCM, which is \(2\sqrt x \sqrt{36-x}\)

Answer 13

that means that i get
\[\sqrt{36-x} + \sqrt{x} = 0\]

Answer 14

now multiply by the conjugate.

Answer 15

Actually, it's a bit weird... both of them are technically positive so...

Answer 16

I think you forgot a negative in your derivative.

Answer 17

Nevermind, your derivative is correct.

Answer 18

find the common denominator, and solve for x.

Answer 19

solving for y with the conjugates got me x = 18 and y = 18 which is the answer for part b

Answer 20

let me check...

Answer 21

Actually, I don't think \(x=18\) is a critical number.

Answer 22

\(\LARGE\color{black}{\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{36-x}}=0 }\)

\(\LARGE\color{black}{\sqrt{x}\sqrt{36-x}=0 }\)
critical numbers 0 or 36.

Answer 23

(I have a prayer right now, excuse me... will see you:))

Answer 24

okay no problem.

Answer 25

The only critical numbers are \(x=0\) and \(x=36\), since it is undefined at these points.

Also, since they're the interval boundaries.

Answer 26

so then i just plug them into the equation and see what it is right?

Answer 27

I 've got to go, but i think i get this. Thanks