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math125-lecture 1.4

# math125-lecture 1.4 - Sec 1.4 Thursday 8:41 PM Section1dot4...

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Unformatted text preview: Sec 1.4 Thursday, February 04, 2010 8:41 PM Section1dot4 Page 1 A function, where every variable in the expression defining the function is raised to the first power (at most) EX: f ( x , y ) = 2 x - (5/7) y 1.4.1 Thursday, February 04, 2010 8:41 PM Section1dot4 Page 2 A function, where every variable in the expression defining the function is raised to the first power (at most) EX: f ( x , y ) = 2 x - (5/7) y R = 10 x + 10 y solve for y to get the equation of a line y = - x + R/10 Thus, for any value of R , the line has the same slope m = -1 1.4.2 Thursday, February 04, 2010 8:41 PM Section1dot4 Page 3 A function, where every variable in the expression defining the function is raised to the first power (at most) EX: f ( x , y ) = 2 x - (5/7) y R = 10 x + 10 y solve for y to get the equation of a line y = - x + R/10 Thus, for any value of R , the line has the same slope m = -1 R = 500, y = - x + 500/10 = - x + 50 R = 1000, y = - x + 100 R = 1500, y = - x + 150 R = 2000, y = - x + 200 R = 2500, y = - x + 250 100 200 1.4.3 Thursday, February 04, 2010 8:41 PM Section1dot4 Page 4 A function, where every variable in the expression defining the function is raised to the first power (at most) EX: f ( x , y ) = 2 x - (5/7) y R = 10 x + 10 y solve for y to get the equation of a line y = - x + R/10 Thus, for any value of R , the line has the same slope m = -1 R = 500, y = - x + 500/10 = - x + 50 R = 1000, y = - x + 100 R = 1500, y = - x + 150 R = 2000, y = - x + 200 R = 2500, y = - x + 250 100 200 1.4.4 Thursday, February 04, 2010 8:41 PM Section1dot4 Page 5 A function, where every variable in the expression defining the function is raised to the first power (at most) EX: f ( x , y ) = 2 x - (5/7) y R = 10 x + 10 y solve for y to get the equation of a line y = - x + R/10 Thus, for any value of R , the line has the same slope m = -1 R = 500, y = - x + 500/10 = - x + 50 R = 1000, y = - x + 100 R = 1500, y = - x + 150 R = 2000, y = - x + 200 R = 2500, y = - x + 250 We pick the line with the largest revenue that still intersects the feasible region Here the line for R = 2000 has one point in common with the feasible region, i.e. (0, 200). Thus, max revenue is 10(0) + 10(200) = \$2000 100 200 1.4.5 Thursday, February 04, 2010 8:41 PM Section1dot4 Page 6 Assume that we have a linear function f ( x , y ) = ax + by Lines of constancy are lines along which the value of f ( x , y ) is constant It is the set of values ( x , y ) for which f ( x , y ) = c for some constant...
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math125-lecture 1.4 - Sec 1.4 Thursday 8:41 PM Section1dot4...

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