hey, ive got this problem, which isnt that hard, but i cannot see how to get to the solution. Any help will be very much appreciated!!
Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1
its a lagrange multiplier, help!!
hey, ive got this problem, which isnt that hard, but i cannot see how to get to the solution. Any help will be very much appreciated!!
Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1
its a lagrange multiplier, help!!
We have the condition
$\displaystyle x^2 + 4y^2 = 1$
So
$\displaystyle x^2 + 4y^2 - 1 = 0$
Thus we may consider the maximum of the function
$\displaystyle f(x, y) = 4xy = 4xy + \lambda(x^2 + 4y^2 - 1)$
where $\displaystyle \lambda$ is an arbitrary constant.
Now:
$\displaystyle \frac{\partial f}{\partial x} = 4y + \lambda \cdot 2x = 0$
$\displaystyle \frac{\partial f}{\partial y} = 4x + \lambda \cdot 8y = 0$
$\displaystyle \frac{\partial f}{\partial \lambda} = x^2 + 4y^2 - 1 = 0$
The first equation says
$\displaystyle \lambda = - \frac{2y}{x}$
So the second equation now says:
$\displaystyle 4x + \left ( - \frac{2y}{x} \right ) \cdot 8y = 0$
$\displaystyle 4x - \frac{16y^2}{x} = 0$
$\displaystyle x^2 - 4y^2 = 0$
So we have two equations in two unknowns:
$\displaystyle x^2 - 4y^2 = 0$
$\displaystyle x^2 + 4y^2 - 1 = 0$
Adding the two equations gives
$\displaystyle 2x^2 - 1 = 0$
$\displaystyle x = \pm \frac{\sqrt{2}}{2}$
Thus
$\displaystyle \left ( \pm \frac{\sqrt{2}}{2} \right ) ^2 - 4y^2 = 0$
$\displaystyle 4y^2 = \frac{1}{2}$
$\displaystyle y = \pm \frac{\sqrt{2}}{4}$
So there are a total of 4 solution points:
$\displaystyle \left ( \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{4} \right )$
where the $\displaystyle \pm$ signs are independent.
I leave it to you to determine which are relative maxima or relative minima.
-Dan
Well, there is a more complicated way using second derivatives, but two simple methods also work:
1) Graph it
2) Can you find any values of the function greater than 1?
As I get the values $\displaystyle \pm 1$ when I put in the values of x and y that were derived for possibilities, I would say that
$\displaystyle \left ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{4} \right )$ and $\displaystyle \left ( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{4} \right )$ generate maximum values and that $\displaystyle \left ( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{4} \right )$ and $\displaystyle \left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{4} \right )$ generate the minimum values.
-Dan
We need to solve,
$\displaystyle \nabla g = k\nabla f$ where $\displaystyle f=x^2-y$.
Thus,
$\displaystyle \left< 2x+2y+2, 4y+2x+3 \right> = k\left< 2x , - 1 \right>$
Thus,
$\displaystyle \left\{ \begin{array}{c}2x+2y+2 = 2xk \\ 4y+2x+3 = -k \\ x^2 - y = 1 \end{array} \right\}$
$\displaystyle 2x+2y+2 = 2x(-4y-2x-3)$Originally Posted by The Greatest
$\displaystyle 2x+2y+2=-8xy-4x^2-6x$
$\displaystyle 8xy+4x^2+8x+2y+2=0$
$\displaystyle 4xy+2x^2+4x+y+1=0$
$\displaystyle 4x(x^2-1)+2x^2+4x+(x^2-1)+1=0$
$\displaystyle 4x^3 - 4x + 2x^2 + 4x + x^2 =0$
$\displaystyle 4x^3 + 3x^2 = 0$
$\displaystyle x^2(4x+3)=0$
You are big boy you can do it from here.
Plug each (x, y) point into your g(x, y) function. (The thing, after all, you are trying to minimize.) The two possibilities you have are the only ones where the function is either at a relative maximum or a relative minimum. I see that
$\displaystyle g(0, -1) = 0^2 + 2(-1)^2 + 2 \cdot 0 \cdot (-1) +2 \cdot 0 + 3(-1) = 2 - 3 = -1$
$\displaystyle g \left ( -\frac{3}{5}, -\frac{16}{25} \right ) = ~ ... ~ = -\frac{733}{625}$
Since $\displaystyle -1 > -\frac{733}{625}$ the $\displaystyle
\left ( -\frac{3}{5}, -\frac{16}{25} \right )$ point is your minimum.
-Dan
What we have shown is that the curves defined by the objective and the
constraint kiss at the points we have found. A kissing point may be the
analog of a local maxima, minima or point of inflection . We still need to
classify the point, or at least justify our conclusion that one is a minimum.
In this case we can conclude that the function does have a minima as when
x goes of to infinity the objective becomes arbitrily large, also that any
minima for this objective and constraint must be smooth, so we then know
that one of the kissing points must be a minima, and we have identified which
it is.
RonL
RonL
i was just wondering where this came from, as i thought that the other values were (-0.75,-7/16) which gives a value of -1,27/128. this is smaller than -1 so is the minimum value right?
also to test the minimum using the idea that as
x goes of to infinity the objective becomes arbitrily large. can you actually use this concept to test for a minimum value?!