Question

Calc III Help
Lagrange Multipliers
Maximize: f(x,y)= sqrt(6-x^2-y^2)
Constraint: x+y-2=0

Answer 1

first step is to get the system of equations
got that yet?

Answer 2

The first partials?

Answer 3

get the gradient of f and set the components equal to a constant multiple of the gradient of the constraint
then solve for the constant, lambda

Answer 4

I got the gradient f = -1,-1 is that correct?
Also when I was solving for lambda I am getting -1 and it that doesnt seem right.

Answer 5

how did you get a number for the gradient?
what is the partial of f wrt x ?

Answer 6

Fx = -x / sqrt(6-x^2-y^2)

Answer 7

Isn't the grad F equal to the coeficients in what you are trying to min/max?

Answer 8

no the grad is just \[\nabla f(x,y)=<f_x,f_y>\]left as variables

Answer 9

Alright. So after finding grad f. I am to find the partial for the constraint correct?

Answer 10

partials**

Answer 11

right, you need\[\nabla f=\lambda\nabla g\]where g is the constrain
then set each component equal\[f_x=\lambda g_x\]\[f_y=\lambda g_y\]then solve the system

Answer 12

both Gx and Gy are 1 correct?

Answer 13

right

Answer 14

So I am then just setting the Fx=Fy and solving?

Answer 15

yep

Answer 16

Alright and once I am able to get either everything to equal x or y I would pulg that back into either Fx or Fy which ever is easier to solve?

Answer 17

Or into the constrain?

Answer 18

constraint*

Answer 19

no, you plug it into the constraint
what result did you get from your system?

Answer 20

\[y = (x \sqrt{6-x ^{2}-y ^{2}})\div \sqrt{6-x ^{2}-y ^{2}}\]

Answer 21

I fear you have over-complicated this...

Answer 22

Damn

Answer 23

\[f_x=fy\]\[\frac{-x}{\sqrt{6-x^2-y^2}}=\frac{-y}{\sqrt{6-x^2-y^2}}\]\[x=y\]

Answer 24

Alright. So after putting that back into the constraint x=1 and y=1 correct?

Answer 25

right

Answer 26

Thank you very much for clearing that up. So from the start find Fx, Fy, Gx, and Gy then set Fx= Lambda Gx and Fy= Lambda Gy then solve for lambda in 1 of the 2 equations plug it into the other solve for either x or y then plug that number into the constraint then solve for x and y in the contraint and that will be your max or min.

Answer 27

yes
there could, of course, be three variables at times
also note that lambda always drops out of the problem, so you never actually need to find its value

also, don't forget to check that whatever point you have actually is a min or max
i.e. check the actual value(s) of f you get from the plug-in(s)

you're welcome, and good luck!

Answer 28

my hobby-horse reference for much of this kind of thing is this page

http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

Answer 29

Where I check the max/min I would plug those into the f(x) correct?

Answer 30

exactly

Answer 31

Gotchya. Alright thank you again for the help.

Answer 32

anytime!

Answer 33

Quick question when doing extremas does D=FxxFyy-(Fxy)^2

Answer 34

yes it does :D