maximization problem

**saskadimova** · March 4th 2009, 05:18 PM

Please help on this one....
Don't really understand the problem. What they ask me to do.
It says..

Solve the following maximization problem:

max xy^(1\2)

x,y

subject to X^(2) +(c*y)^2 <= 1,x,y => 0, where

c> 0 .
Let V(c) denote the maximized value of the objective function (as a function of c).

Calculate V ' (c).

**Scott H** · March 5th 2009, 04:48 AM

As always, we must check all critical points in the domain. For a differentiable function, the critical points are the boundary points and the points inside at which $\nabla z=0$ . In our case,

$ \begin{aligned} z&=xy^{\frac{1}{2}}\\ \nabla z&=y^{\frac{1}{2}}\textbf i\,+\frac{1}{2}xy^{-\frac{1}{2}}\textbf j,\\ \end{aligned} $

so the only time $\nabla z$ ever vanishes is when $y=0$ (and thus $z=0$ ) at a boundary point, leaving us with only boundary points for the maximum. Our domain is an elliptical region in the first quadrant bounded by the $x$ and $y$ axes. When $x$ or $y$ equals $0$ , $z=xy^{\frac{1}{2}}=0$ , leaving us with the boundary points at $g(x,y)=x^2+(cy)^2=1$ . By Lagrange's Method, we know that if $(x,y)$ is a maximum, then

$\begin{aligned} \nabla z&=\lambda\nabla g\\ {y}^{\frac{1}{2}}\textbf i\,+\frac{1}{2}{x}{y}^{-\frac{1}{2}}\textbf j\,&=\lambda(2x\textbf i\,+2c^2y\textbf j). \end{aligned} $

Solving this vector equation for $x$ and $y$ will give you a point at which $z$ attains its maximum.

**saskadimova** · March 5th 2009, 08:39 AM

Hey Scott, thx for the help. So if i solve your equation what i get eventually for V(c) is:
c=

Code:

(1\2)^1\2 *x\y

and finally i receive that the first derivative of c- , which is the maximized value of the objective function is :
c' =

Code:

2x * 1\(2^1\2 * y) +2y * 1\(2^1\2 * x)

Do you agree ? Am i right.

**Scott H** · March 5th 2009, 07:23 PM

We can begin with

$\begin{aligned} y^{\frac{1}{2}}&=2\lambda x\\ \frac{y^{\frac{1}{2}}}{2\lambda}&=x \end{aligned} $

and substitute for $x$ :

$\begin{aligned} \frac{1}{2}xy^{-\frac{1}{2}}&=2\lambda c^2y\\ \frac{1}{2}\left(\frac{y^{\frac{1}{2}}}{2\lambda}\ right)y^{-\frac{1}{2}}&=2\lambda c^2y\\ \frac{1}{4\lambda}&=2\lambda c^2y\\ \frac{1}{8\lambda^2c^2}&=y. \end{aligned}$

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