Thread: Determine the minimizers and maximizers of f subject to the given constraints

f(x,y) = 3*(x^3) + 2*(y^3) subject to x^2 + y^2 = 4. This should be done using Lagrange multipliers, but i couldn't solve it.

$\displaystyle \nabla f(x,y)=9x^2\mathbf{i}+6y^2\mathbf{j}$

$\displaystyle \lambda\nabla g(x,y)=2x\lambda\mathbf{i}+2y\lambda\mathbf{j}$

$\displaystyle \displaystyle 9x^2=2x\lambda\Rightarrow \frac{9x}{2}=\lambda$

$\displaystyle \displaystyle 6y^2=2y\lambda\Rightarrow 6y^2=2y\left(\frac{9x}{2}\right)$

$\displaystyle \displaystyle 6y=9x\Rightarrow x=\frac{2}{3}y$

$\displaystyle \displaystyle x^2+y^2=4\Rightarrow \left(\frac{2}{3}y\right)^2+y^2=4$

$\displaystyle \displaystyle \frac{4y^2}{9}+y^2=\frac{13y^2}{9}=4$

$\displaystyle \displaystyle 13y^2=36\Rightarrow y=\pm\frac{6}{\sqrt{13}}$

$\displaystyle \displaystyle x=\left(\frac{1}{3}\right)\left(\frac{\pm 6}{\sqrt{13}}\right)=\pm\frac{2}{\sqrt{13}}$

$\displaystyle \displaystyle f\left(\frac{2}{\sqrt{13}},\frac{6}{\sqrt{13}}\rig ht)=\cdots$

$\displaystyle \displaystyle f\left(\frac{-2}{\sqrt{13}},\frac{-6}{\sqrt{13}}\right)=\cdots$

Originally Posted by enrique

f(x,y) = 3*(x^3) + 2*(y^3) subject to x^2 + y^2 = 4. This should be done using Lagrange multipliers, but i couldn't solve it.

Could you please show what you tried? I see nothing at all unusual about this problem.

For any "Lagrange multiplier" problem in two variables, you will get two equations of the form $\displaystyle g(x,y)= \lambda f(x,y)$ and $\displaystyle h(x,y)= \lambda k(x,y)$.

Dividing one equation by the other gives an equation that does not involve $\displaystyle \lambda$: $\displaystyle \frac{g(x,y)}{h(x,y)}= \frac{f(x,y)}{k(x,y)}$ which, in this case, is particularly simple.

This is one form of Lagrange multipliers, that is taught in calculus courses. This question is from the optimization book of Nash & Sofer, they use some different notation and solve similar types of questions using matrices, which is confusing for me. But i see that this type of problems can be solved by just using Calculus "Lagrange multipliers", right?