Question

Lagrange Multiplier:
f(x,y) = 5x^2 + 2y^2
xy=3

Answer 1

image included

Answer 2

i don't understand why it isn't 399/10

i got
x=\[(2\sqrt{3} )/ 5\]
y=\[(5\sqrt{3} )/ 2\]

Answer 3

which multiplied together = 3

so then you would plug x and y back into f(x,y) = 399/10

Answer 4

it's really x = above, y = above
and x = - <(negative)> above, y = - <(negative)> above
but either one with come out with the same minimum value

Answer 5

399/10 =39.9

what if we just use x=1 and y=3

then x*y=3

and

\[f(x,y) = 5x^2 + 2y^2\]
\[f(1,3) = 5\cdot 1^2 + 2\cdot 3^2=5+18=23\]

so you don't have the min

Answer 6

looks like the min is \(6\sqrt{10}\)

Answer 7

hrm but you can't just plug numbers in like that right??
because you have to solve the system?

Answer 8

I'm just showing you that your answer is not correct

Answer 9

oh

Answer 10

so you can do test cases like that?

Answer 11

Just by observation I could tell that your answer did not give the min

Answer 12

why don't you show your work and we will tell you where your mistake is

Answer 13

yeah just kinda hard to put it into the computer

Answer 14

\[\nabla f(x,y,z)=\lambda\nabla g(x,y,z)\]I'll try to see ho far I can go before I mess up, I'm not always great with these

Answer 15

yeah, this stuff isn't really that hard conceptually but i have big time trouble with finding all the solutions to systems of equations

Answer 16

k so it will come out with
10x=\[\lambda\]y
4x=\[\lambda\]x
xy=3

Answer 17

er

Answer 18

\[\nabla f=<10x,4y>\]\[\lambda\nabla g=<\lambda x,\lambda y>\]\[10x=\lambda x\implies\lambda=10\]\[4y=\lambda y\implies\lambda=4\]ok I seem to be confused...

Answer 19

x= 3/y
plugging into first:
10(3/y) = z * y
z = 30/y^2
plugging both into second:
y^3 = 22.5
y = 2.823

Answer 20

x=0 y=0 ?

Answer 21

obviously not...

Answer 22

oh you went for lambda first

Answer 23

\[10x=\lambda y\]
\[10x^2=\lambda xy=3\lambda \]

\[4y=\lambda x\]
\[4y^2=\lambda xy=3\lambda\]

\[\Rightarrow 10x^2=4y^2\]

Answer 24

you don't need to know lambda

Answer 25

I know that, but I couldn't eliminate it

Answer 26

yeah, so just eliminate it first

Answer 27

I was just demonstrating that I was lost

Answer 28

oh I messed up the system, duh!

Answer 29

yep...that would cause a problem ;)

Answer 30

ah k so
sqrt 10 * x = 2 * y

Answer 31

so then you use that with the last problem
so basically just get rid of lambda then use the resulting equations
10x^2 - 2*y = 0
x*y = 3

Answer 32

\[\sqrt{10}x=\pm2y\]

Answer 33

\[y=\pm\frac{\sqrt{10}}{2}x\]

Answer 34

thank you both sooo much

Answer 35

yeah, I'm not going to restate Zarkon, I'm tired anyway
g'night y'all

Answer 36

night