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anonymous
 one year ago
Given the linear programming problem, use the method of corners to determine where the minimum occurs and give the minimum value
minimize
C=x+3y
Subjectto
x less than equal to 3
y less than equal to 5
x greater then equal to 0
y greater to equal than 0
anonymous
 one year ago
Given the linear programming problem, use the method of corners to determine where the minimum occurs and give the minimum value minimize C=x+3y Subjectto x less than equal to 3 y less than equal to 5 x greater then equal to 0 y greater to equal than 0

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im not sure how to ... I can read a graph already drawn but creating one myself is a different story.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let's start with this.dw:1441475021450:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(y \ge 0\) is everything above xaxisdw:1441475235023:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and \(x \ge 0\) is everything right of the yaxis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then for \(y \le 3\) shade everything below the line \(y=3\)dw:1441475376673:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(x \le 5\) is left of the line x = 5dw:1441475429193:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So your corners are the corners of the black rectangle above, where all 4 inequalities are valid

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Write the coordinates of those 4 corners then test them all in the constraint \(C=x+3y\) to find the smallest

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would 0,3 and 5,0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, those are two of the corners. (0, 0) and (5, 3) are the others
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