Exercise 16

There is a useful interpretation of the Lagrange multiplier $\lambda$. Assume that we want to optimize a function $f$ with constraint $g(x,y)=c$. Recall that an optimal solution occurs at a point $(x_0, y_0)$ where $\nabla f = \lambda \nabla g$. As the constraint changes, so does the point at which the optimal solution occurs. So we can think of the optimal point as a function of the parameter $c$, that is $x_0 = x_0(c)$ and $y_0=y_0(c)$. The optimal value of $f$ subject to the constraint can then be considered as a function of $c$ defined by $f(x_0(c), y_0(c))$. The Chain Rule shows that $$\begin{gathered} \frac{df}{dc} = \frac{\partial f}{\partial x_0} \frac{dx_0}{dc} + \frac{\partial f}{\partial y_0} \frac{dy_0}{dc}. \end{gathered}$$

1. Use the fact that $\nabla f = \lambda \nabla g$ at $(x_0,y_0)$ to explain why $$\begin{gathered} \frac{df}{dc} = \lambda \frac{dg}{dc}. \end{gathered}$$

2. Use the fact that $g(x,y) = c$ to show that $$\begin{gathered} \frac{df}{dc} = \lambda. \end{gathered}$$ Conclude that $\lambda$ tells us the rate of change of the function $f$ as the parameter $c$ increases. In other words, the value of $\lambda$ is approximately how much the optimal value of the function $f$ will change if we increase the value of $c$ by 1 unit.

3. In this example, we did not explicity calculate the value of $\lambda$ for our maximum value of the objective function. Use our work in this example to find the value of $\lambda$ and explain in context of our Preview Activity what the value $\lambda$ tells us about the package.

4. Suppose that the maximum value of a function $f = f(x,y)$ subject to a constraint $g(x,y) = 100$ is $236$. When using the method of Lagrange multipliers and solving $\nabla f = \lambda \nabla g$, we obtain a value of $\lambda = 15$ at this maximum. Find an approximation to the maximum value of $f$ subject to the constraint $g(x,y) = 98$.