# SOLUTION: Maximize z = 2x + 4y subject to the constraints x &#8805; 0, y &#8805; 0, x + y &#8805; 1, 3x + 2y <= 6

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 Question 251750: Maximize z = 2x + 4y subject to the constraints x ≥ 0, y ≥ 0, x + y ≥ 1, 3x + 2y <= 6 Answer by Edwin McCravy(13211)   (Show Source): You can put this solution on YOUR website! Maximize z = 2x + 4y subject to the constraints x ≥ 0, y ≥ 0, x + y ≥ 1, 3x + 2y <= 6 ``` When x is "greater than" it means "to the right of". That means that the feasible region is on or to the right of the line whose equation is which is the line which is the y-axis. When y is "greater than" it means "above" That means that the feasible region is on or above the line whose equation is which is the line which is the x-axis. Those two mean the feasible region is in this upper right- hand part of the xy-coordinate system: If you solve that for x you get When x is "greater than" it means "to the right of" If you solve that for y you get When y is "greater than" it means "above". That means that the feasible region is on or above and to the right of the line whose equation is . The x-intercept of that line is (1,0), and the y-intercept is (0,1). That's above and to the right of the green line below, and of course right of the y-axis and above the x-axis: If you solve that for x you get When x is "less than" it means "to the left of" If you solve that for y you get When y is "less than" it means "below". That means that the feasible region is on or bolow and to the left of the line whose equation is . The x-intercept of that line is (2,0), and the y-intercept is (0,3). That's below and to the left of the blue line below, and of course right of the y-axis and above the x-axis, and above and to the right of the green line: The corner points of that region are (1,0), (2,0), (0,3), and (0,1) So we list them in this table with the objective function evaluated at each corner point: Corner point | Value of z = 2x+4y ----------------|--------------------------- (1,0) | z = 2(1)+4(0) = 2+0 = 2 | (2,0) | z = 2(2)+4(0) = 4+0 = 4 | (0,3) | z = 2(0)+4(3) = 0+12 = 12 | (0,1) | z = 2(0)+4(1) = 0+4 = 4 The instructions were to maximize the objective function. So we see that the maximum value of the objective function is 12 when x=0 and y=3. [Note: If the problem had asked us to minimize the objective function. the answer would have been that the minimum value of the objective function would be 2 when x=1 and y=0.] Edwin```