Investigate function for conditional extremums

[lavreniukkk]

Good day, can you help to investigate function $$z=x^3-3xy^2+18y$$ for conditional extremums, if $$3x^2y-y^3-6x=0$$I've stuck during solving of system of equations based on a Lagrange function. I'm not sure if it has solutions.

 

[topsquark]

Can you show us how you got the Lagrange equations?

-Dan

 

[lavreniukkk]

Can you show us how you got the Lagrange equations?

-Dan

Sure, $$F(x,y,λ)=x^3-3xy^2+18y+λ(3x^2y-y^3-6x)$$

 

[JeffM]

Sure, $$F(x,y,λ)=x^3-3xy^2+18y+λ(3x^2y-y^3-6x)$$

That LaGrangian function is correct for this problem although it might be better to show

$$F(x) = x^3 - 3xy^2 + 18y + \lambda(\{3x^2y - y^3 - 6x\} - 0) \implies \\ F(x) =x^3 - 3xy^2 + 18y + \lambda(3x^2y - y^3 - 6x) .................................edited$$
But it does not answer topsquark’s question at all.

What you calll F(x), I usually call L.

The next step, as indicated by topsquark’s question, is to compute the partial derivatives of the LaGrangian function. Note that you now have three independent variables, namely x, y, and $\lambda.$ So you will get three partials. The third step is to equate each partial to zero. That gives you three equations for three unknowns. You can combine these two steps as follows

$$\dfrac{\delta L}{\delta \lambda} = 0 \implies 3x^2 - y^3 - 6x = 0$$.
What are the rest of the results from equating the partial derivatives to zero?