Introduction to Lagrange Multipliers

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Problem Set 7’s estimation problem.

Problem 2 in this problem set.

The key idea is to estimate the function’s derivatives from the data in the table. Then you can use those derivatives and a function value from a table entry near (2.1, 3.05) to do linear estimation. The derivative with respect to x can be estimated by looking at how the function’s value changes across rows, and the derivative with respect to y by looking at changes up or down columns.

Colloquium

By Madeline Klein, Geneseo Class of 2015 and Hannover Reassurance of America

“This is What Actuaries Do”

Thursday, October 27 2:30 - 3:20 pm

ISC 131

Lagrange Multipliers

Based on section 3.8 in the textbook.

Key Ideas

To solve an optimization problem with constraint g(x,y) = 0 and objective function f(x,y), use the fact that minimums or maximums occur where…

• ∇f(x,y) = λ ∇g(x,y)
• g(x,y) = 0

Examples

Suppose the production function I used yesterday to motivate optimization has a constraint that x + y ≤ 10. So we have

• z = 2 x^0.4 y^0.5
• x + y ≤ 10

What values of x and y maximize production, and what is the maximum amount of product you can produce?

To find the x and y values, find the gradients and set up the equations from the “Key Ideas”:

To solve the equations, notice that the first 2 imply that 0.8 y^0.5 / x^0.6 = x^0.4 / y^0.5. Use that equation to express x in terms of y, then plug that into the constraint (the third equation) to find the actual value of x. Once you have a value for x, use it to solve for y:

Finally, plug these values of x and y into a calculator or Mathematica to find the corresponding z: z ≈ 8.56.

Next

Another example of optimization with Lagrange multipliers.

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