# Lagrange Multipliers

• March 11th 2013, 11:20 AM
apatite
Lagrange Multipliers
Use the method of lagrange multipliers to determine the maximum value of f(x,y)= x^a y^b (a and b are two fixed positive constants) subject to constraint x + y= 1. (assume x,y bigger than zero)

Help would be appreciated! I get ax^a-1 = bY^b-1 but dont know how to do the rest. I cant isolate x based on only a and b.
• March 11th 2013, 03:23 PM
HallsofIvy
Re: Lagrange Multipliers
Don't forget your constraint: $x+ y= 1$ so that $y= 1- x$. Putting that into $ax^{a-1} = by^{b-1}$ gives $ax^{a-1}= b(1- x)^{b-1}$. Of course, that is a polynomial of degree a- 1 or b-1, whichever is larger, and there is no simple way of solving that.
• March 12th 2013, 05:57 AM
majamin
Re: Lagrange Multipliers
Quote:

Originally Posted by apatite
Use the method of lagrange multipliers to determine the maximum value of f(x,y)= x^a y^b (a and b are two fixed positive constants) subject to constraint x + y= 1. (assume x,y bigger than zero)

Help would be appreciated! I get ax^a-1 = bY^b-1 but dont know how to do the rest. I cant isolate x based on only a and b.

Let $g(x,y)=x+y$

$\nabla f=-\lambda\nabla g$ or:
$\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)=-\lambda\left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)$

$\frac{\partial f}{\partial x} = ax^{a-1}y^b$
$\frac{\partial f}{\partial y} = bx^ay^{b-1}$
$\frac{\partial g}{\partial x} = \frac{\partial g}{\partial y} = 1$

So: $\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)=-\lambda\left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) \implies$

$(ax^{a-1}y^b,bx^ay^{b-1})=-\lambda(1,1)=(-\lambda,-\lambda)$
Using the fact that $g(x,y)=1$ or $y= 1 - x$ then
$ax^{a-1}(1-x)^b=-\lambda$
$bx^a(1-x)^{b-1}=-\lambda$

$\frac{ax^{a-1}(1-x)^b}{bx^a(1-x)^{b-1}}=1$

$\frac{a(1-x)}{bx}=1$

Or: $x= \frac{a}{a+b}$ so that $y= \frac{b}{a+b}$ i.e. f(x,y) is a maximum when x and y take on these values to get

$f_{max} = \left(\frac{a}{a+b}\right)^a \left( \frac{b}{a+b} \right)^b$

As a final step, you must check concavity here to verify that f is a maximum at these values.