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notes-unconstrained-max

# notes-unconstrained-max - OPMT 5701 Two Variable...

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OPMT 5701 Two Variable Optimization Using Calculus For Maximization Problems One Variable Case If we have the following function y = 10 x x 2 we have an example of a dome shaped function. To fi nd the maximum of the dome, we simply need to fi nd the point where the slope of the dome is zero, or dy dx = 10 2 x = 0 10 = 2 x x = 5 and y = 25 Two Variable Case Suppose we want to maximize the following function z = f ( x, y ) = 10 x + 10 y + xy x 2 y 2 Note that there are two unknowns that must be solved for: x and y . This function is an example of a three-dimensional dome. (i.e. the roof of BC Place ) To solve this maximization problem we use partial derivatives. We take a partial derivative for each of the unknown choice variables and set them equal to zero z x = f x = 10 + y 2 x = 0 The slope in the ”x” direction = 0 z y = f y = 10 + x 2 y = 0 The slope in the ”y” direction = 0 This gives us a set of equations, one equation for each of the unknown variables. When you have the same number of independent equations as unknowns, you can solve for each of the unknowns. rewrite each equation as y = 2 x 10 x = 2 y 10 substitute one into the other x = 2(2 x 10) 10 x = 4 x 30 3 x = 30 1

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x = 10 similarly, y = 10 REMEMBER: To maximize (minimize) a function of many variables you use the technique of partial di ff erentiation. This produces a set of equations, one equation for each of the unknowns. You then solve the set of equations simulaneously to derive solutions for each of the unknowns. Second order Conditions (second derivative Test) To test for a maximum or minimum we need to check the second partial derivatives. Since we have two fi rst partial derivative equations ( f x , f y ) and two variable in each equation, we will get four second partials ( f xx , f yy , f xy , f yx ) Using our original fi rst order equations and taking the partial derivatives for each of them (a second time) yields: f x = 10 + y 2 x = 0 f y = 10 + x 2 y = 0 f xx = 2 f yy = 2 f xy = 1 f yx = 1 The two partials, f xx , and f yy are the direct e ff ects of of a small change in x and y on the respective slopes in in the x and y direction. The partials, f xy and f yx are the indirect e ff ects, or the cross e ff ects of one variable on the slope in the other variable’s direction. For both Maximums and Minimums , the direct e ff ects must outweigh the cross e ff ects Rules for two variable Maximums and Minimums 1. Maximum f xx < 0 f yy < 0 f yy f xx f xy f yx > 0 2. Minimum f xx > 0 f yy > 0 f yy f xx f xy f yx > 0 3. Otherwise, we have a Saddle Point From our second order conditions, above, f xx = 2 < 0 f yy = 2 < 0 f xy = 1 f yx = 1 and f yy f xx f xy f yx = ( 2)( 2) (1)(1) = 3 > 0 therefore we have a maximum.
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