Answer:
(3, 0)
Maximum Value of Objective Function = 15
Explanation:
This is a problem related to Linear Programming(LP)
In linear programming, the objective is to maximize or minimize an objective function subject to a set of constraints.
For example, you may wish to maximize your profits from a mix of production of two or more products subject to resource constraints.
Or, you may wish to minimize cost of production of those products subject to resource constraints..
The given LP problem can be stated in standard form as
Max 5x - 4y
s.t.
-x + 3 ≥ y
y ≤ 0.5x + 1
x ≥ 0, y ≥ 0
The last two constraints always apply to LP problems which means the decision variables x and y cannot be negative
It is standard to express these constraints with the decision variables on the LHS and the constant on the RHS
Rewriting the above LP problem using standard notation,
Let's rewrite the constraints using the standard form:
- x + 3 ≥ y
→ -x - y ≥ -3
→ x + y ≤ 3 [1]
y ≤ 0.5x + 1
→ -0.5x + y ≤ 1 [2]
The LP problem becomes
Max 5x - 4y
s. t.
x + y ≤ 3 [1]
-0.5x + y ≤ 1 [2]
x ≥ 0 [3]
y ≥0 [4]
With an LP problem of more than 2 variables, we can use a process known as the Simplex Method to solve the problem
In the case of 2 variables, it is possible to solve analytically or graphically. The graphical process is more understandable so I will use the graphical method to arrive at the solution
The feasible region is the region that satisfies all four constraints shown.
The graph with the four constraint line equations is attached. The feasible region is the dark shaded area ABCD
The feasible region has 4 corner points(A, B,C, D) whose coordinates can be computed by converting each of the inequalities to equalities and solving for each pair of equations.
It can be proved mathematically that the maximum of the objective function occurs at one of the corner points.
Looking at [1] and [2] we get the equalities
x + y = 3 [3]
-0.5x + y = 1 [4]
Solving this pair of equations gives x = 4/3 and y = 5/3 or (4/3, 5/3)
Solving y = 0 and x + y = 3 gives point x = 3, y =0 (3,0)
The other points are solved similarly, I will leave it up to you to solve them
The four corner points are
A(0,0)
B(0,1)
C(4/3, 5/3)
D(3,0)
The objective function is 5x - 4y
To find the values of x and y that maximize the objective function,
plug in each of the x, y values of the corner points
Ignoring A(0,0)
we get the values of the objective function at the corner points as
For B(0,1) => 5(0) - 4(1) = -4
For C(4/3, 5/3) => 5(4/3) - 4(5/3) = 20/3 - 20/3 = 0
For D(3, 0) => 5(3) - 4(0) = 15
So the values of x and y which maximize the objective function are x = 3 and y = 0 or point D(3,0)