using Lagrange multiplier to find min and max points

Question

I need to find a minimum and maximum values of the function $f(x,y)=3πxy^2$ with the following constraint: $6+6πxy+3πy^2=12$.

I just seem to get stuck in the middle as I have too many variables. Thank you in advance.

Answer 1

$$ \nabla f=3\pi(y^2,2xy) $$

$$ g(x,y)=6\pi x y+3\pi y^2-6 $$

$$ \nabla g=3\pi(2y,2x+2y) $$

$$ \nabla f = \lambda\nabla g $$

Note that $\lambda \neq 0$, as if it were zero $\nabla f=0$ which implies $y=0$ and $g(x,0)\neq 0$.

$$ y^2=2\lambda y $$

$$ 2xy=\lambda(2x+2y) $$

$$ \frac{2xy}{y^2}=\frac{\lambda(2x+2y)}{2\lambda y}=1+\frac{x}{y} $$

$$ \frac{2x}{y}=1+\frac{x}{y} $$

which implies $x=y$.

now the constraint must be satisfied:

$$ g(x,x)=6\pi x^2+3\pi x^2-6=0 $$

$$ x=\pm\sqrt{\frac{6}{9\pi}}=\pm\sqrt{\frac{2}{3\pi}} $$

so we have two extrema on the constraint

$$ (x,y)=(\sqrt{\frac{2}{3\pi}},\sqrt{\frac{2}{3\pi}}) $$

$$ (x,y)=(-\sqrt{\frac{2}{3\pi}},-\sqrt{\frac{2}{3\pi}}) $$

If you plug these into $f(x,y)$ you will see that the positive solution gives a maximum and the negative value a minimum.

Answer 2

$6+6\pi xy+3\pi y^2=12 \Rightarrow \pi y^2+2\pi xy=2.$

$L=f(x,y)+\lambda(2-\pi y^2-2\pi xy)$

$L_x=3\pi y^2-2\pi \lambda y=0 \stackrel{y \ne0}\Rightarrow \lambda = \frac{3y}{2}$

$L_y=6\pi xy-2\pi \lambda y-2\pi x \lambda=0 \Rightarrow 3xy-3y^2=0 \stackrel{y\ne0} \Rightarrow x=y$

$L_{\lambda}=2-\pi y^2-2\pi xy=0 \Rightarrow x=\pm \sqrt{\frac{2}{3\pi}} =y.$

Can you handle the rest?

Answer 3

solving your condition for $x$ you will get $$x=\frac{2-\pi y^2}{2\pi y}$$ plugging this in your function you will get $$f\left(\frac{2-\pi y^2}{2\pi y},y\right)=\frac{3}{2}y(2-\pi y^2)$$ and you have to solve $$3-\frac{9}{2}\pi y^2=0$$