maximize Z = 30x+50y
subject to
2x+4y=100 and x+5y=80
I assume you are supposed to do a graphical solution.
first, change each constraint equation to slope/inercept form.
y=-1/2 x+25 and
y=-1/5 x + 16
Now, on an x/y graph, plot those lines. Now what is neat about this solution, the max or min is always at an intersection.
Here, if you assume x>0, and y>0, then you have 4 corners (intersections).
0,0
16,0
0,25 and finally where the lines cross at 30,10
check all those points, I did them in my head.
Then, take your objective function Z, and calculate the value at each of these corners: One will be the max solution.
Now to make a genius of yourself, try a point x,y inside the area, and somewhere along any line, just to prove to yourself the maximum and minimum will always be on a corner.
To solve this linear programming problem, we can use the Simplex Method.
Step 1: Convert the constraints to equalities by introducing slack variables:
2x + 4y + s1 = 100 (Constraint 1)
x + 5y + s2 = 80 (Constraint 2)
Step 2: Convert the objective function to maximize form by multiplying it by -1:
-Z = -30x - 50y
Let's set up the initial simplex table:
____________________________________________________
| Basic Variables | x | y | s1 | s2 | RHS |
|-----------------------|---------|---------|---------|---------|-----------|
| Z = 0 | -30 | -50 | 0 | 0 | 0 |
| s1 = 100 | 2 | 4 | 1 | 0 | 100 |
| s2 = 80 | 1 | 5 | 0 | 1 | 80 |
Step 3: Choose the most negative coefficient in the Z row as the pivot column. In this case, the most negative coefficient is -50 in the y column.
Step 4: Select the pivot row. Divide the RHS values by the coefficients in the pivot column. In this case, the minimum ratio is for s2 with 16.
Step 5: Perform row operations to make the pivot element (1 in row 3, column 2) equal to 1 and other elements in the pivot column equal to 0.
Row 3 ÷ 5 → Row 3 (Divide row 3 by the pivot element 5)
____________________________________________________
| Basic Variables | x | y | s1 | s2 | RHS |
|-----------------------|---------|---------|---------|---------|-----------|
| Z = 80 | -4 | 0 | 0 | 10 | 400 |
| s1 = 60 | 1/5 | 0 | 1/5 | -1/25 | 20 |
| y = 16 | 1/5 | 1 | -1/5 | 1/25 | 16 |
Step 6: Update the Z equation by subtracting the coefficients of the pivot row multiplied by the corresponding coefficients in the Z row.
Row 1 + 4 * Row 3 → Row 1 (Multiply row 3 by 4, then add it to row 1)
Row 2 - 1/5 * Row 3 → Row 2 (Multiply row 3 by -1/5, then subtract it from row 2)
____________________________________________________
| Basic Variables | x | y | s1 | s2 | RHS |
|-----------------------|---------|---------|---------|---------|-----------|
| Z = 80 | 0 | 0 | 2/5 | 99/25 | 464 |
| s1 = 60 | 0 | 0 | 1/5 | -6/25 | 4 |
| y = 16 | 1/5 | 1 | -1/5 | 1/25 | 16 |
Step 7: Repeat steps 3 to 6 until all coefficients in the Z row are non-negative.
Step 8: As all coefficients in the Z row are non-negative, the current solution is optimal.
Z = 80 when x = 0, y = 16, s1 = 60, s2 = 0.
So, the optimum solution is:
x = 0
y = 16
Z = 80, where Z is maximized.
To solve this linear programming problem and maximize the objective function Z = 30x + 50y, subject to the constraints 2x + 4y = 100 and x + 5y = 80, we can use the method of substitution or the method of elimination. Here, we will use the method of substitution.
Step 1: Solve one of the constraints for a variable.
Let's solve the second equation, x + 5y = 80, for x:
x = 80 - 5y
Step 2: Substitute the expression for x in terms of y into the other constraint.
Substituting x into the first constraint, we get:
2(80 - 5y) + 4y = 100
Step 3: Simplify and Solve for y.
Expanding and simplifying the equation, we have:
160 - 10y + 4y = 100
-6y = 100 - 160
-6y = -60
Dividing both sides by -6:
y = (-60) / (-6)
y = 10
Step 4: Substitute the value of y back into one of the original constraints to find the corresponding value of x.
Substituting y into the second constraint, we have:
x + 5(10) = 80
x + 50 = 80
x = 80 - 50
x = 30
Step 5: Calculate the value of the objective function.
Substituting the values of x and y into the objective function, we have:
Z = 30(30) + 50(10)
Z = 900 + 500
Z = 1400
Therefore, the maximum value of Z is 1400, which occurs when x = 30 and y = 10.