When the constraint … is a strict inequality … at the minimum…

These problems allow inequality constraints, optional but good.

When the constraint $h\ge k$ is a strict inequality $h>k$ at the minimum, the multiplier is $\lambda =0$. Explain the reasoning: For a small increase in $k$, the same minimizer is still available (since $h>k$ leaves room to move). Therefore $f_{\min }$ is (changed) (unchanged), and $\lambda =d{f}_{\min }/dk=$______.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §13.7 Constraints and Lagrange Multipliers, #28