Consider the two images. I understand how the equality constraint $\lambda$ can be positive and negative because the gradient can point to or from the minimum but always a scaler $\lambda$ away from the gradient of the function (at the minimum). I also understand that inequality constraints $\mu$ can be either active or non-active and when it is active, it is considered an equality constraint i.ie equal to zero.
My question is then why can't $\mu$ be negative too, what if the inequality constraint looks like the figure 10.7 then why cant it have an opposite direction to the gradient.
I found a similar question but it did not help my confusion.
As you mentioned that the KKT multiplier for an equality constraint is clear to you, let me discuss about the KKT multiplier for inequalities. However, still note that in Figure 10.7 you have to walk along the equality constraint, whereas in Figure 10.9 you have to search in a region.
In Figure 10.9, you are solving a problem of the following kind:
\begin{array}[t]{l} \min f(x)\\ s.t.\\ \quad c(x) \ge 0\\ \end{array}
The gradient to the constraint is pointing inwards. Going in the direction of the constraint gradient you will find feasible region, however, your function value will deteriorate. Going in the opposite direction your function value will improve but there is no feasible region. Going along the constraint boundary will not lead to any improvement in function value (locally). Therefore, the point is a local optimum. At this point the gradients of the objective function and the constraint are aligned, or in other words: $$\nabla{f(x^{\ast})} = \mu \nabla{c(x^{\ast})}, \mu \ge 0$$
Now consider a situation where $\mu$ is negative. In this case you are talking about a situation where the two gradients are in opposite directions. Moving along the gradient of the constraint will keep you in feasible region, and your function value will also improve (for minimization problems). Therefore, a negative $\mu$ will not represent an optimal scenario.