# Solve the Objective Function Subject to Restrictions

Solve for the values of x and y which maximize and minimize the following objective function:

Q = 3x + 2y

Subject to the following restrictions:

y ≥ (3/2) x - 3

y ≤ -x + 7

x ≥ 0

y ≥ 0

### Comments for Solve the Objective Function Subject to Restrictions

 Jul 04, 2012 Solve Objective Function Subject to Restrictions by: Staff Plot the restrictions (click the link below to view the plot): y ≥ (3/2) x - 3 (the red line and all the area above the red line) y ≤ -x + 7 (the green line and all the area below the green line) x ≥ 0 (all the area to the right of the vertical blue line) y ≥ 0 (all the area above the horizontal pink line) Bounded Feasible Region (shaded yellow area) When these boundaries are plotted, they form the Bounded Feasible Region (click link to view, use the Backspace key to return to this page): http://www.solving-math-problems.com/images/graph-objective-funciton-boundaries-2012-07-04.png The bounded feasible region (shown as the yellow shaded area) defines a maximum value and a minimum value for the objective function (Q = 3x + 2y). Corner points: The corner points are shown in the following graph (click link to view, use the Backspace key to return to this page): http://www.solving-math-problems.com/images/graph-feasible-region-corner-points-2012-07-04.png The corner points are: (0,0), (0,7), (4,3), (2,0) Find the maximum and minimum values of Q by computing the values of Q at the corner points. Q = 3x + 2y Corner point: (0,0) Q = 3*0 + 2*0 = 0, minimum value of Q Corner point: (0,7) Q = 3*0 + 2*7 = 14 Corner point: (4,3) Q = 3*4 + 2*3 = 18, maximum value of Q Corner point: (2,0) Q = 3*2 + 2*0 = 6 Maximum value of Q is: 18; when x = 4 and y = 3 Minimum value of Q is: 0; when x = 0 and y = 0 Thanks for writing. Staff www.solving-math-problems.com