Question

Because we use the envelope theorem in constrained optimization problems often in the text, proving this theorem in a simple case may help develop some intuition. Thus, suppose we wish to maximize a function of two variables and that the value of this function also depends on a parameter, \(a: f\left(x_{1}, x_{2}, a\right) .\) This maximization problem is subject to a constraint that can be written as: \(g\left(x_{1}, x_{2}, a\right)=0\) a. Write out the Lagrangian expression and the first-order conditions for this problem. b. Sum the two first-order conditions involving the \(x\) 's. c. Now differentiate the above sum with respect to \(a-\) this shows how the \(x\) 's must change as a changes while requiring that the first-order conditions continue to hold. A. As we showed in the chapter, both the objective function and the constraint in this problem can be stated as functions of \(a: f\left(x_{1}(a), x_{2}(a), a\right), g\left(x_{1}(a), x_{2}(a), a\right)=0 .\) Dif ferentiate the first of these with respect to \(a\). This shows how the value of the objective changes as \(a\) changes while kecping the \(x^{\prime}\) s at their optimal values. You should have terms that involve the \(x\) 's and a single term in \partialflaa. e. Now differentiate the constraint as formulated in part (d) with respect to \(a\). You should have terms in the \(x^{\prime}\) 's and a single term in \(\partial g / \partial a\) f. Multiply the results from part (e) by \(\lambda\) (the Lagrange multiplier), and use this together with the first-order conditions from part (c) to substitute into the derivative from part (d). You should be able to show that \\[ \frac{d f\left(x_{1}(a), x_{2}(a), a\right)}{d a}=\frac{\partial f}{\partial a}+\lambda \frac{\partial g}{\partial a^{\prime}} \\] which is just the partial derivative of the Lagrangian expression when all the \(x^{\prime}\) 's are at their optimal values. This proves the enyclope theorem. Explain intuitively how the various parts of this proof impose the condition that the \(x\) 's are constantly being adjusted to be at their optimal values. 8\. Return to Example 2.8 and explain how the cnvelope theorem can be applied to changes in the fence perimeter \(P-\) that is, how do changes in \(P\) affect the size of the area that can be fenced? Show that, in this case, the envelope theorem illustrates how the Lagrange multiplier puts a value on the constraint.

Short answer

Answer: The envelope theorem is used for analyzing how the optimal value of an objective function changes with respect to a parameter when subject to a constraint. It simplifies calculations by only considering the direct effect of the parameter on the objective function and the constraint, rather than accounting for each variable's indirect effects.

Step-by-step solution

a. Write the Lagrangian expression and the first-order conditions

Let the Lagrangian function be defined as: \\[ L\left(x_1, x_2, a, \lambda \right) = f\left(x_1, x_2, a\right) - \lambda g\left(x_1, x_2, a\right) \\] The first-order conditions require us to take the partial derivatives of the Lagrangian function with respect to all its variables and set them equal to zero: 1. \\[ \frac{\partial L}{\partial x_1} = \frac{\partial f}{\partial x_1} - \lambda \frac{\partial g}{\partial x_1} = 0 \\] 2. \\[ \frac{\partial L}{\partial x_2} = \frac{\partial f}{\partial x_2} - \lambda \frac{\partial g}{\partial x_2} = 0 \\] 3. \\[ \frac{\partial L}{\partial \lambda} = g\left(x_1, x_2, a\right) = 0 \\]

b. Sum the two first-order conditions involving the x's

Add the first two first-order conditions obtained in step a: \\[ \frac{\partial f}{\partial x_1} + \frac{\partial f}{\partial x_2} = \lambda \left( \frac{\partial g}{\partial x_1} + \frac{\partial g}{\partial x_2} \right) \\]

c. Differentiate the sum with respect to a

Differentiate both sides of the equation with respect to a: \\[ \frac{\partial^2 f}{\partial x_1 \partial a}+\frac{\partial^2 f}{\partial x_2 \partial a}=\lambda\left(\frac{\partial^2 g}{\partial x_1 \partial a}+\frac{\partial^2 g}{\partial x_2 \partial a}\right) \\]

d. Differentiate the objective function and constraint with respect to a

Differentiate the objective function: \\[ \frac{d f(x_1(a), x_2(a), a)}{d a} = \frac{\partial f}{\partial x_1}\frac{d x_1}{d a} + \frac{\partial f}{\partial x_2}\frac{d x_2}{d a} + \frac{\partial f}{\partial a} \\] Differentiate the constraint: \\[ \frac{d g(x_1(a), x_2(a), a)}{d a} = \frac{\partial g}{\partial x_1}\frac{d x_1}{d a} + \frac{\partial g}{\partial x_2}\frac{d x_2}{d a} + \frac{\partial g}{\partial a} \\]

e. Multiply the result of part (e) by 𝜆 and substitute into the derivative from part (d)

First, multiply the result from part (e) by 𝜆: \\[ \lambda \frac{d g(x_1(a), x_2(a), a)}{d a} = \lambda \frac{\partial g}{\partial x_1}\frac{d x_1}{d a} + \lambda \frac{\partial g}{\partial x_2}\frac{d x_2}{d a} + \lambda \frac{\partial g}{\partial a} \\] Now, substitute the expressions from the first-order conditions (step a) involving x's: \\[ \frac{d f(x_1(a), x_2(a), a)}{d a} = \left(\frac{\partial f}{\partial x_1}\right)\left(\lambda \frac{\partial g}{\partial x_1}\right) + \left(\frac{\partial f}{\partial x_2}\right)\left(\lambda \frac{\partial g}{\partial x_2}\right) + \frac{\partial f}{\partial a} \\] Now, we can simplify to obtain: \\[ \frac{d f(x_1(a), x_2(a), a)}{d a} = \frac{\partial f}{\partial a} + \lambda \frac{\partial g}{\partial a} \\] This proves the envelope theorem. The proof intuitively shows that by constantly differentiating the objective function and the constraint with respect to the parameter a, we are adjusting the x's to their optimal values so that we reach the maximum value of the objective function subject to the constraint.