Question

Find A and B given that the function y=Ax^(-1/2)+Bx^(1/2)
has a minimum value of 6 at x = 9

Answer 1

So you found y' correct?

Answer 2

No I found \[B = (A/-9)+2\]

Answer 3

Use \[
y(6)=9
\]and \[
y'(6)=0
\]Those two equations will allow you to solve for the variables.

Answer 4

But after that I tried to plug in and then i didnt work out

Answer 5

:O I didnt think about the derivative

Answer 6

But wio didnt you get those backwards because x=9 and has a value of 6

Answer 7

So its f(9)=6
and f'(9)=0

Answer 8

You are right matt.

Answer 9

Lol cool

Answer 10

Ok so I solved for B and I pluged it all in and got this

\[y=\frac{ A }{ \sqrt(x) }+(\frac{ -A }{ 9 }+2)\sqrt(x)\]

Answer 11

How did you get B was equal to that?

Answer 12

6=A(9)^(-1/2)+B(9)^(1/2)
Since the sqrt(9)=3 It turns into this
6=(A/3)+B(3)
Then I subtracted B(3) to the left side
6-B(3)=(A/3)
Then I subtracted 6 on the right side
-B(3)=(A/3)-6
Then I divided -3 on both sides to get
B=(-A/9)+2

Answer 13

Oh I left off the 3 in front of the B. Looks good.

Answer 14

So we have one linear equation from the y(9)=6.
now let's look at y'(9)=0.

Answer 15

I would just leave in terms of A and B until the end. But you can do this like you are.

Answer 16

Just start off by taking the derivative?

Answer 17

yep.

Answer 18

Oh well then lol

Answer 19

A and B are just constants.
\[(x^\frac{1}{2})'=? \text{ and } (x^\frac{-1}{2})'=?\]

Answer 20

ok

Answer 21

For some reason im at this point

\[1=\frac{ A }{ 2x^3}-\frac{ A }{ 18 }\]

But if I try to

Answer 22

I dont know what im doing

Answer 23

oh so did you keep it in terms of A and B.

Answer 24

It doesn't matter. I just think it looks less confusing if we keep in terms of A and B but we can do in terms of A if you so wish.

Answer 25

No its from where I left off but I guess ill try to start off with in terms of A or B because its confusing

Answer 26

Is this what you got for y'?
\[y'=\frac{-1}{2}A \frac{1}{x^\frac{3}{2}}+\frac{1}{2}(\frac{-A}{9}+2)\frac{1}{x^\frac{1}{2}}\]

Answer 27

Did you do y'(9)=0?

Answer 28

set x equal to 9 and y'=0?

Answer 29

Ok

Answer 30

Ok now I got A=9

Answer 31

So then B=1

Answer 32

Correct?

Answer 33

Yes. that looks good. :)

Answer 34

Thanks a bunch myininaya your always a huge help! :D