Solve the following linear programming problem: Maximize Z = 8X + 7Y, such that 3X + 2Y ≤ 6000, 2X + Y ≤ 4500, and X, Y ≥ 0.

Steps to Solve

1. Identify the Constraints The inequalities that define the feasible region are:

   • ( 3X + 2Y \leq 6000 )
   • ( 2X + Y \leq 4500 )
   • ( X, Y \geq 0 )

2. Graph the Constraints To graph the inequalities, we convert them to equations:

   • From ( 3X + 2Y = 6000 ), isolate ( Y ): $$ Y = 3000 - \frac{3}{2}X $$
   • From ( 2X + Y = 4500 ), isolate ( Y ): $$ Y = 4500 - 2X $$ Plot these lines on a graph to find the feasible region.

3. Find the Intersections of the Constraints Solve for intersections of the equations to find corner points:

   • Set ( 3X + 2Y = 6000 ) and ( 2X + Y = 4500 ).
   • Substitute ( Y = 4500 - 2X ) into ( 3X + 2(4500 - 2X) = 6000 ): $$ 3X + 9000 - 4X = 6000 $$ $$ -X + 9000 = 6000 $$ $$ X = 3000 $$
   • Substitute back to get ( Y: Y = 4500 - 2(3000) = -1500 ) (Not valid since ( Y \geq 0 ))

   Examine other intersections:

   • Solve ( 3X + 2Y = 6000 ) and ( Y = 0 ) to find ( (2000, 0) ).
   • Solve ( 2X + Y = 4500 ) and ( Y = 0 ) to find ( (2250, 0) ).
   • Solve ( 3X + 2Y = 6000 ) and examine other constraints to find valid ( (0, 3000) ) and ( (0, 4500) ).

4. Evaluate the Objective Function at Each Corner Point Calculate ( Z ) at the feasible corner points:

   • At ( (2000, 0) ): $$ Z = 8(2000) + 7(0) = 16000 $$
   • At ( (2250, 0) ): $$ Z = 8(2250) + 7(0) = 18000 $$
   • At ( (0, 3000) ): $$ Z = 8(0) + 7(3000) = 21000 $$
   • At ( (0, 4500) ): $$ Z = 8(0) + 7(4500) = 31500 $$

5. Determine the Maximum Value of Z Compare the values obtained:

   • ( 16000, 18000, 21000, 31500 ) The maximum value occurs at ( (0, 4500) ).