How do you minimize and maximize #f(x,y)=x^3-y# constrained to #x-y=4#?

we'll do it first as a problem in single variable calculus.

it's all the usual stuff from here on

i'll do a Lagrange Multiplier next to compare. the basic premise is that with

or

I'm asking for a second opinion on this next bit.

Because there is no simple way to explore the nature of the turning points, especially with more complex problems, when using the LM approach. You can often play with the physical reality and reason a solution but there is no quick second derivative check, sadly, that i am aware of.

would be of any use here.

To minimize and maximize $f(x, y) = x^3 - y$ constrained to $x - y = 4$, follow these steps:

1. Solve the constraint equation $x - y = 4$ for $y$ to get $y = x - 4$.
2. Substitute $y = x - 4$ into the objective function $f(x, y) = x^3 - y$ to get the function in terms of $x$ only: $g(x) = x^3 - (x - 4)$.
3. Find the critical points of $g(x)$ by setting its derivative equal to zero and solving for $x$.
4. Determine the corresponding $y$-values for each critical point using the constraint equation.
5. Evaluate the objective function $f(x, y)$ at each critical point to find the maximum and minimum values.

Alternatively, you can use Lagrange multipliers to solve this constrained optimization problem.