# maximization problem

• Mar 4th 2009, 05:18 PM
maximization problem

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).
• Mar 5th 2009, 04:48 AM
Scott H
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.
• Mar 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. (Wondering)
• Mar 5th 2009, 07:23 PM
Scott H
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}