1. Lagrange Multipliers

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

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

Constraint:

$\displaystyle x+y=1$

Using the Lagrange Multiplier method, I managed to obtain

$\displaystyle 3x^2=\lambda$

$\displaystyle -2y=\lambda$

Solving for the variables gave me:

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

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

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

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

Then I end up with the quadratic, after simplifying;

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

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

Thank you

2. Re: Lagrange Multipliers

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 $\displaystyle y=1-x$, we obtain

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

So, $\displaystyle \varphi'(x)=3x^2-2x+2>0\quad \forall x\in\mathbb{R}$ (there are no extrema).

3. Re: Lagrange Multipliers

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