# Very important, Lagrange multiplier

1. Apr 17, 2012

### odporko

Guys, i would be really greatfull if someone help me with this because i really dont know how to deal with this math problem: Find the maximum and minimum values of f = x^(1/4) + y^(1/3) on the boundary of g = 4*x+ 6*y = 720.

2. Apr 17, 2012

### Ray Vickson

You need to first show what you have done before we are allowed to offer help.

RGV

3. Apr 17, 2012

### odporko

L ( x , y ) = x^(1/4) + y^(1/3) + λ ( 4*x + 6*y − 720 )

I am not able to partialy derivate this equation, it always has weird numbers and after that i am not able to continue because i dont understand L. multiplier very well.

4. Apr 17, 2012

### Ray Vickson

Are you saying you do not know how to differentiate x^(1/4) and x with respect x? If that is really true you need to go back and review introductory calculus.

RGV

5. Apr 18, 2012

### odporko

1. The problem statement, all variables and given/known data

f(x,y) = x^(1/4)*y^(1/3), g(x,y) = 4*x +6*y=720
sorry i had a mistake before

2. Relevant equations

∇f=λ∇g
y^(1/3)/(4*x^(3/4))= 4*λ
x^(1/4)/(3*y^(2/3))= 6*λ

these are partial derivations, and we want to express y just with lambda and here is something that really terrifies me: 4096*((104976*y^(8/3)*λ)^(9/4))*λ, am i thinking right? or am i going wrong?

6. Apr 18, 2012

### odporko

it goes for y=(1/ 812 479 653 374 328*λ^(13/4))^(1/5)... and now my mind already blew up

7. Apr 18, 2012

### odporko

4*(1/450868486864896*λ^(11/3))^(1/5)+ 6*(1/ 812479653374328*λ^(13/4))^(1/5)-720=0
well this is the equation after i get x and y and now i am pretty sure i did something wrong...

8. Apr 18, 2012

### HallsofIvy

Staff Emeritus
Good. So eliminate $\lambda$ by dividing the first equation by the second:
$$\frac{y^{1/3}}{4x^{3/4}}\frac{3y^{2/3}}{x^{1/4}}=\frac{3}{4}\frac{y}{x}= \frac{2}{3}$$
so we must have y= (8/9)x. Put that into the equation of the constraint, 4x+6y= 720, to solve for x.

Well, what we want to do is to solve three equations in x, y, and $\lambda$ for x and y. We do not necessarily "want to express y just with lambda". There are many ways of solving for x and y. Since finding a value for $\lambda$ is not part of the solution, I suggest eliminating $\lambda$ first by dividing one equation by another.

9. Apr 18, 2012

### Ray Vickson

You can use the first equation to solve for y in terms of (x,λ), then substitute that y into the second equation to get an equation containing only x and λ. Solve for x in terms of λ, then substitute that x into the expression for y. You will end up with x and y expressed only in terms of λ, and the expressions are not too horrible:
$$x = \frac{2^{4/5}\, 3^{2/5}}{2304\, \lambda^{12/5}}, \; y = \frac{2^{4/5}\, 3^{2/5}}{2592\, \lambda^{12/5}}.$$
Substituting these into the equation for g you get a simple linear equation in the variable
$\lambda^{12/5},$ so you can solve it and pick the relevant real, positive root. This turns out to be
$$\lambda = \frac{5^{7/12}\, 3^{1/12}\, 7^{5/12}\, 2^{1/6}}{720} \doteq 0.009827792083.$$ Now you can find x and y:
$$x = 540/7, \; y = 480/7, \; f(x,y) = \frac{540^{1/4}\, 7^{5/12} \,480^{1/3}}{7} \doteq 12.13030337.$$ This is the constrained max.

The constrained min cannot be found by Lagrange multipliers. Just recognize that f(x,y) is not defined for x < 0 and y < 0, so we are restricted to the region {x ≥ 0, y ≥ 0}, and that f = 0 when x = 0 or y = 0. Thus, there are two constrained min points, at (x,y) = (0,720/6) and (x,y) = (720/4,0), both giving f = 0.

RGV

10. Apr 19, 2012

### odporko

Thank you very much, you really really helped me probably pass the most important subject this semester :)

11. Apr 19, 2012

### NewtonianAlch

An entire subject just on Lagrange Multipliers? Lol.