# SOLUTION: How do I minimize: z=13x+9y subject to: 3x+4y>=21; 8x+4y>=32; x>=0; y>=0 ?

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 Click here to see ALL problems on Linear-systems Question 468975: How do I minimize: z=13x+9y subject to: 3x+4y>=21; 8x+4y>=32; x>=0; y>=0 ?Answer by ccs2011(207)   (Show Source): You can put this solution on YOUR website!First graph your bounds which are defined as: 3x+4y>=21; 8x+4y>=32; x>=0; y>=0 Each of these represents an area above a line. Determine slope and y_intercept of each line by rewriting in slope-intercept form. y = mx + b x = 0 and y =0 are just the axis of the graph Rewrite 3x+4y>=21 by solving for y: Subtract 3x on both sides Divide by 4 on both sides Rewrite 8x+4y>=32 by solving for y: Subtract 8x on both sides Divide by 4 on both sides Now graph these lines All the possible (x,y) points are in the region that is on or above both lines Lets look at z=13x+9y The goal is to minimize z, there is a direct variation with z to x and y so to decrease z you must decrease x and/or y. If you pick any point in the possible region and move it straight down until you hit one of the 2 lines, then you kept the same x but decreased y thus decreasing z. From that logic the smallest z values will be on the line. Typically the smallest z values will be on the endpoints. There are 3 end-points, x-intercept (7,0) y-intercept (0,8) and point where lines intersect. To find where they intersect, set 2 equations equal to each other. Eliminate fractions by multiplying everything by 4 Add 8x on both sides Subtract 21 on both sides Divide by 5 on both sides Substitute back into equation to find y Finally evaluate z at each end-point, smallest value will be minimum For (7,0): z = 13*7 = 91 For (0,8): z = 9*8 = 72 For (11/5,18/5): z = 13*(11/5) + 9*(18/5) = 305/5 = 61 Therefore z has minimum of 61 at point(11/5,18/5)