Question

Find the absolute maximum and minimum of f(x, y) = 2x+3y subject to the constraint squareroot(x) + squareroot(y) = 5

Answer 1

Use Lagrange multipliers

Answer 2

Do you know how to find ReverseTriangle.F and ReverseTriangle.G?

Answer 3

No i do not.

Answer 4

Have you learned about Lagrange Multipliers lol?

Answer 5

how can u ask this questions then xD

Answer 6

Use f(x, y) = 2x+3y to find ReverseTriangle.F

Answer 7

Then use squareroot(x) + squareroot(y) = 5 to find ReverseTriangle.G

Answer 8

do u want to see a great video on lagrange multipliers?

Answer 9

very useful to know

Answer 10

I have learned lagrange multipliers however I can only seem to find one critical point (9,4). The answer I have tells me that the minimum is (9.4) but the maximum is (0,25). How would you find the point (0,35)? I have not heard of this reverse triangle stuff

Answer 11

he just means delta F and delta G

Answer 12

ohhh i see.

Answer 13

well I still cannot find point (0,25)

Answer 14

as a critical point.

Answer 15

Find ReverseTriangle.F and ReverseTriangle.G first! Then we'll tell you how to go from there :)

Answer 16

gradient f = <2, 3> gradient g = <1/(2squareroot(x)),1/(2squareroot(y))>

Answer 17

Solving everything down gives me x=3/2root(y) which I plug in to g to give me x=9
Then pluging it back gives me the point (9.4)

Answer 18

How do I get (0,25) which is the maximum?

Answer 19

Ok so you have your two ReverseTriangles! Now we are getting somewhere.
Now you want to make each component in your ReverseTriangle.F as a relationship with your ReverseTriangle.G times a Scalar

Answer 20

\[\Delta F = \lambda \Delta G\]

Answer 21

I have done that but solving the equation for x and substituting it into g gives me only the point (9,4)

Answer 22

The key here is that I have no Idea how to get the point (0,25) as a critical point which the answer tells me is the maximum.

Answer 23

Put in (9,4) back into function F!

Answer 24

It gives me 30

Answer 25

Are you sure you solved for lambda properly?

Answer 26

lambda is equal to 12

Answer 27

\[<2,3> = \lambda <\frac{ 1 }{ 2\sqrt{x} },\frac{ 1 }{ 2\sqrt{y} }>\]

Answer 28

yes I have that.

Answer 29

\[2=\frac{ \lambda }{ 2\sqrt{x} }, 3=\frac{ \lambda }{ 2\sqrt{y} }\]

Answer 30

\[4\sqrt{x}=6\sqrt{y}\]

Answer 31

Yes I have all of that. The one problem is that using all of this, I am unable to find the point (0,25) as a critical point when it clearly is one.

Answer 32

What does Y equal in terms of X?

Answer 33

(4/9)x

Answer 34

squareroot(x) + squareroot(\[\sqrt{x}+\sqrt{\frac{ 4x }{ 9 }}=5\]) = 5

Answer 35

\[=\sqrt{x}+\frac{ \sqrt{4x} }{ \sqrt{9} }=\]

Answer 36

5

Answer 37

\[\sqrt{x}+\frac{ 2\sqrt{x} }{ 3 }=5\]

Answer 38

\[=3\sqrt{x}+2\sqrt{x}=15\]

Answer 39

\[=5\sqrt{x}=15\]

Answer 40

..lmao..

Answer 41

..I see where this is heading..give me a sec

Answer 42

I have all of this. The end result is x=9 and plugging that into (4/9)x=y gives y=4. The question is how I can find out (0,25) is a critical point.

Answer 43

Is (25,0) also a critical point?

Answer 44

According to the answers, (25,0) is the maximum while the (9,4) i've already found is the minimum.

Answer 45

(0,25) sry

Answer 46

I just need to know how to find out that (0,25) is a critical point.

Answer 47

I'll be back. While I'm gone, let's see if The Big Guns, AKA @oldrin.bataku can help.

Answer 48

Try to think back to Calc I.
When we had a function and needed to find critical points, we would just use the derivative (very similar to what we're doing here with the Lagrange thingy-ma-bobber).

But when we had a function `within an interval` we also had to check the end points of that interval.


The same thing is happening here.
So you've found critical points within the region, good.
Now we need to check the boundaries of the region as well.

We can do that by letting one of the variables be 0.

\[\large x=0 \qquad \rightarrow \qquad \sqrt{y}=5 \qquad \rightarrow \qquad y=25\]

Plugging these values into our original function gives us,\[\large f(0,25)=2\cdot0+3\cdot25\]


It would also make sense to check the other boundary I guess,

\[\large y=0\]



Understand how that works?

Answer 49

Ah... I had always assumed that lagrange would find the critical points on the boundaries as well. Thanks for your help!

Answer 50

@FutureMathProfessor \(\nabla=\text{\nabla}\)

Answer 51

\[\nabla\]

Answer 52

W00T!