(Sensitivity) Certainly … has

These problems develop the theory of Lagrange multipliers.

(Sensitivity) Certainly $L=f-\lambda (g-k)$ has $\partial L/\partial k=\lambda$. Since $L=f_{\min }$ and $g=k$ at the minimum point, this seems to prove the key formula $d{f}_{\min }/dk=\lambda$. But $x_{\min },y_{\min },\lambda$, and $f_{\min }$ all change with $k$. We need the _total_ derivative of $L(x,y,\lambda ,k)$:
$$\frac{dL}{dk}=\frac{\partial L}{\partial x}\frac{dx}{dk}+\frac{\partial L}{\partial y}\frac{dy}{dk}+\frac{\partial L}{\partial \lambda }\frac{d\lambda }{dk}+\frac{\partial L}{\partial k}\frac{dk}{dk}.$$
Equation (1) at the minimum point should now yield the sensitivity formula $d{f}_{\min }/dk=\lambda$.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §13.7 Constraints and Lagrange Multipliers, #15