Lagrange Multipliers

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May 15th 2012, 11:34 AM
iPod

Lagrange Multipliers

Find the maxima and the minima of the given function given the constraint.

$f(x,y)=x^3-y^2$

Constraint:

$x+y=1$

Using the Lagrange Multiplier method, I managed to obtain

$3x^2=\lambda$

$-2y=\lambda$

Solving for the variables gave me:

$x=\pm \sqrt{\frac{\lambda}{3}}$

$y=-\frac{\lambda}{2}$

So then I plug these values into my constraint to end up with

$\pm \sqrt{\frac{\lambda}{3}}-\frac{\lambda}{2}=1$

Then I end up with the quadratic, after simplifying;

$3\lambda^2+8\lambda+12=0$

However, this quadratic can't be solved using real numbers. Have I done anything wrong or am I using the wrong approach?

Thank you
May 16th 2012, 09:29 AM
FernandoRevilla

Re: Lagrange Multipliers

Quote:

Originally Posted by iPod

However, this quadratic can't be solved using real numbers. Have I done anything wrong or am I using the wrong approach?

You are right. There are no critical points. Another way: substituting $y=1-x$ , we obtain

$\varphi (x)=f(x,1-x)=x^3-(1-x)^2=x^3-x^2+2x-1$

So, $\varphi'(x)=3x^2-2x+2>0\quad \forall x\in\mathbb{R}$ (there are no extrema).
May 16th 2012, 11:23 AM
HallsofIvy

Re: Lagrange Multipliers

Because a specific value $\lambda$ is NOT a part of the solution, I find it simplest to eliminate $\lambda$ first. Here, instead of solving for x and y in terms of $\lambda$ , I would use $3x^2= \lambda$ and $-2y= \lambda$ to say that $3x^2= -2y$ so that $y= -(3/2)x^2$ . Put that into the condition x+ y= 1 to get $x- (3/2)x^2= 1$ which is equivalent to $3x^2- 2x+ 2= 0$ . The discriminant for that quadratic equation is $(-2)^2- 4(3)(2)= 4- 24= -20$ so, yes, there is no real value of x satisfying this problem.