A rectangular box without a lid is to be made from 50m² of cardboard. Find the maximum volume of such a box.( i know how to solve this in the conventional way, i am trying to figure out how to do it using this method, where the problem needs to be converted to a minimisation method then using the Karush-Kuhn-Tucker conditions)
objective function and constraints will be: $\max [f(x,y,z)]=xyz$ subject to $h(x,y,z)=xy+2yz+2zx-50=0$
Now to Convert to standard Lagrangian minimisation problem I use the fact that $\min f(x)=\max -f(x)$ then the langrangian $$L(x,y,z)=f(x,y,z)-kc(x,y,z)= -xyz-k(xy+2yz+2zx-50)$$ and this should now be a minimisation problem
The Karush-Kuhn-Tucker conditions state that for a minimisation problem: $\nabla L(x^*,y^*,z^*,k^*)=0$ and $c(x^*,y^*,z^*)=0$ where the values of $x^*,y^* z^*$ are the solutions for the minimum.
then if i can show that the hessian matrix for the langrangian $L(x^*,y^*,z^*,k^*)$ is positive definite then this shows that the values i have found corespond to the global minimum.
I cant actually solve the equations $\nabla L(x*,y*,z*,k*)=0$ and $c(x*,y*,z*)=0$. When i put them in matrix form i get a mess, so this is what i am stuck on and i am starting to get a bit confused.