I would like to solve the following optimization problem: \begin{array}{ll} \text{maximize} & 52 - 4(x-1)^{2/3} - x^2 \\ \text{subject to}& x\ge 0 \\ &x \le 11 \end{array}
I know that the answer is $x=1$, a point in which the derivative of the function is not defined.
How could I find that using traditional Lagrangean multipliers? This is how I started:
$$L(x,\lambda_{1}, \lambda_{2}) = 52 - 4(x-1)^{2/3} - x^2 + \lambda_{1}x+\lambda_{2}(11-x)$$
Which yields the following first order conditions:
\begin{align} -\frac83(x-1)^{-1/3} - 2x + \lambda_{1} - \lambda_{2} = 0 \\ \lambda_{1}x=0 \\ \lambda_{2}(11-x)=0 \\ x \ge 0, x \le 11 \\ \lambda_{1} \ge 0, \lambda_{2} \ge 0 \end{align}
After some calculations I found out that $\lambda_{1}$ has to be $0$. However, if $\lambda_{1} =0$, then $\lambda_{2} = 0$. This leads to:
$$ -\frac83(x-1)^{-1/3} - 2x =0$$
Which has no solution.
What am I missing here?
The short answer is this: people often forget that critical numbers aren't only where the derivative is equal to $0$, but they are also where the derivative is undefined.
The process I advise people to take is this:
1) Are you constrained by intervals? If so, include any interval endpoints as critical numbers.
2) Are you able to find the derivative of your given function? If so, find where the derivative is equal to $0$, and also, if applicable, find where your derivative is undefined.
The set of $x$-values that are gathered from steps 1) and 2) above are your critical numbers.
In this situation, we see that $0$, $11$ are critical numbers immediately, and since the derivative is undefined at $x = 1$, $1$ is also a critical number. Thus, all three of these numbers should be tested. You should input these values into the objective function and compare their outputs directly.
Note that IF the $x$-value $x_0$ is the location of a local extrema AND a function $f$ is differentiable at $x_0$, it is the case that $f^{\prime}(x_0) = 0$. But this DOES NOT mean that if $f^{\prime}(x_0) = 0 \implies $ there is an extrema at $x_0$. Take $f(x) = x^3$ for example.