带约束的多元目标函数极值函数推导及求解方法问询

Hey there! Let's break this down step by step to figure out how to find that extreme value function (M(K, L)) and address your question about the Hessian for 4 variables.

First, deriving (M(K, L)) #

The extreme value function (M(K, L)) is just your original objective function evaluated at the optimal solutions you've already found ((K_1^{}, L_1^{}, K_2^{}, L_2^{})).

First, let's confirm the objective function from your first-order conditions: it looks like you're maximizing (or minimizing) (f(K_1, L_1, K_2, L_2) = \sqrt{K_1} + 2\sqrt{L_1} + 3\sqrt{K_2} + 4\sqrt{L_2}) (since the partial derivatives match this form). If that's correct, substitute your optimal values into this function:

[
\begin{align*}
M(K, L) &= f\left(\frac{K}{10}, \frac{L}{5}, \frac{9K}{10}, \frac{4L}{5}\right) \
&= \sqrt{\frac{K}{10}} + 2\sqrt{\frac{L}{5}} + 3\sqrt{\frac{9K}{10}} + 4\sqrt{\frac{4L}{5}} \
&= \frac{\sqrt{K}}{\sqrt{10}} + \frac{2\sqrt{L}}{\sqrt{5}} + 3\times\frac{3\sqrt{K}}{\sqrt{10}} + 4\times\frac{2\sqrt{L}}{\sqrt{5}} \
&= \frac{\sqrt{K} + 9\sqrt{K}}{\sqrt{10}} + \frac{2\sqrt{L} + 8\sqrt{L}}{\sqrt{5}} \
&= \frac{10\sqrt{K}}{\sqrt{10}} + \frac{10\sqrt{L}}{\sqrt{5}} \
&= \sqrt{10K} + 2\sqrt{5L}
\end{align*}
]

If your objective function is different, just repeat this process: plug (K_1^{}, L_1^{}, K_2^{}, L_2^{}) into whatever your original target function is, and simplify—this gives you (M(K, L)), which tells you the extreme value for any given (K) and (L).

Next, handling the Hessian for constrained optimization with 4 variables #

You're right that a standard Hessian isn't the right tool here—since you have equality constraints, we use a bordered Hessian matrix instead of the unconstrained Hessian. Here's how to approach it:

1. Construct the Lagrangian:
   Your Lagrangian is already implied by your first-order conditions:
   [
   \mathcal{L} = \sqrt{K_1} + 2\sqrt{L_1} + 3\sqrt{K_2} + 4\sqrt{L_2} - \lambda_1(K - K_1 - K_2) - \lambda_2(L - L_1 - L_2)
   ]

2. Build the bordered Hessian:
   The bordered Hessian includes second partial derivatives of the Lagrangian, plus the gradients of your constraint equations as "borders". For your problem (4 variables, 2 constraints), the matrix looks like this:
   [
   H = \begin{bmatrix}
   0 & 0 & -1 & 0 & -1 & 0 \
   0 & 0 & 0 & -1 & 0 & -1 \
   -1 & 0 & \mathcal{L}{K_1K_1} & 0 & \mathcal{L}{K_1K_2} & 0 \
   0 & -1 & 0 & \mathcal{L}{L_1L_1} & 0 & \mathcal{L}{L_1L_2} \
   -1 & 0 & \mathcal{L}{K_2K_1} & 0 & \mathcal{L}{K_2K_2} & 0 \
   0 & -1 & 0 & \mathcal{L}{L_2L_1} & 0 & \mathcal{L}{L_2L_2}
   \end{bmatrix}
   ]
   Where:

   • The top-left 2x2 block is zeros (since we're bordering with constraints)
   • The top-right and bottom-left blocks are the gradients of your constraints ((g_1 = K - K_1 - K_2 = 0) and (g_2 = L - L_1 - L_2 = 0))
   • The bottom-right 4x4 block is the Hessian of the Lagrangian (second partial derivatives)

3. Evaluate second partial derivatives:
   Calculate the second derivatives from your Lagrangian:

   • (\mathcal{L}{K_1K_1} = -\frac{1}{4}K_1^{-3/2}), (\mathcal{L}{L_1L_1} = -\frac{1}{2}L_1^{-3/2})
   • (\mathcal{L}{K_2K_2} = -\frac{9}{4}K_2^{-3/2}), (\mathcal{L}{L_2L_2} = -2L_2^{-3/2})
   • All cross-partials (like (\mathcal{L}_{K_1K_2})) are zero, since the objective function terms don't mix variables.

4. Check determinant signs for extremality:
   For a maximization problem with (m) constraints, we look at the determinants of the bordered Hessian starting from the ((2m+1))-th order (here, 5th order) and higher. For a concave objective function (which yours is—all square-root terms are concave, and sums of concave functions are concave) and linear constraints (which define a convex set), your critical point is automatically a global maximum—you might not even need to compute the bordered Hessian to confirm this!

备注：内容来源于stack exchange，提问作者Waseem Bughio