# Thread: Lagrange Multipliers

1. ## 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.

2. ## Re: Lagrange Multipliers

Don't forget your constraint: $\displaystyle x+ y= 1$ so that $\displaystyle y= 1- x$. Putting that into $\displaystyle ax^{a-1} = by^{b-1}$ gives $\displaystyle ax^{a-1}= b(1- x)^{b-1}$. 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.

3. ## Re: Lagrange Multipliers 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 $\displaystyle g(x,y)=x+y$

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

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

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

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

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

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

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

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

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

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