# Thread: maximization problem

1. ## 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).

2. 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 $\displaystyle \nabla z=0$. In our case,

\displaystyle \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 $\displaystyle \nabla z$ ever vanishes is when $\displaystyle y=0$ (and thus $\displaystyle z=0$) 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 $\displaystyle x$ and $\displaystyle y$ axes. When $\displaystyle x$ or $\displaystyle y$ equals $\displaystyle 0$, $\displaystyle z=xy^{\frac{1}{2}}=0$, leaving us with the boundary points at $\displaystyle g(x,y)=x^2+(cy)^2=1$. By Lagrange's Method, we know that if $\displaystyle (x,y)$ is a maximum, then

\displaystyle \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 $\displaystyle x$ and $\displaystyle y$ will give you a point at which $\displaystyle z$ attains its maximum.

3. 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.

4. We can begin with

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

and substitute for $\displaystyle x$:

\displaystyle \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}