Step 1: Define the given inequalities and graph the feasible region
We have the following inequalities that define our feasible region:
1. \(x + y \leq 30\)
2. \(x + 2y \leq 40\)
3. \(x \leq 25\)
4. \(x \geq 0\)
5. \(y \geq 0\)
Begin by plotting these inequalities on a graph, and then determine the feasible region.
Step 2: Identify the corner points
The feasible region is a polygon determined by the intersection of the given inequalities. The following corner points found are:
- A: (0, 0)
- B: (0, 20)
- C: (20, 10)
- D: (25, 5)
- E: (25, 0)
Step 3: Evaluate the objective function at each corner point
Now evaluate the value of the objective function, \(P = 4x + 5y\), at each corner point:
- A: \(P(0, 0) = 0\)
- B: \(P(0, 20) = 100\)
- C: \(P(20, 10) = 180\)
- D: \(P(25, 5) = 145\)
- E: \(P(25, 0) = 100\)
The optimal solution is found at point C, with \(x = 20\) and \(y = 10\), and the maximum value of the objective function is \(P = 180\).
Step 4: Find the range of values for the coefficient of x
Now we need to find the range of values for the coefficient of x without changing the optimal solution (corner point C). To do this, we will examine the objective function at points B and D, neighboring corner points of the optimal solution.
Let the new coefficient of x be represented by a constant k. The new objective function would be \(P = kx + 5y\). At point B, the new objective function is \(P_B = 5y = 100\), and at point D, the new objective function is \(P_D = kx + 5y = 145\). From \(P_B = P_D\), we get:
\(100 = 145 - k(25)\)
Solve for k to obtain the range of values satisfying this condition.
\(k = \frac{145-100}{25}\)
The range of possible values for the coefficient of x is \((0, \frac{45}{25})\).
Step 5: Find the range of values for resource 1
Resource 1 is represented by the constraint \(x + y \leq 30\). To determine how far the constraint can be changed without changing the optimal solution, we can look at the neighboring constraint \(x + 2y \leq 40\) because it also defines the optimal solution point C. The current value of resource 1 is 30. Let's assume that the new constraint for resource 1 is \(x + y \leq 30 + k\), where k can take positive or negative values. At the optimal solution, \(x = 20\) and \(y=10\).
Now, the constraint \(x + 2y \leq 40\) will change too and must be satisfied:
\(20 + 20 + k \leq 40\)
Solve for k:
\(k \le 0\)
The range of values resource 1 can take is \([30, 30)\).
Step 6: Find the shadow price of resource 1
The shadow price of a resource represents the change in the objective function's value resulting from an incremental increase in the availability of that resource. Since the range of values for resource 1 only allows for a decrease, according to our calculations above, the shadow price for resource 1 is 0.
Step 7: Identify binding and nonbinding constraints
Binding constraints are the ones that directly affect the optimal solution, and nonbinding constraints do not impact it. Since the optimal solution lies at the intersection of constraints 1 (\(x + y \leq 30\)) and 2 (\(x + 2y \leq 40\)), these constraints are binding. Constraints 3 (\(x \leq 25\)), 4 (\(x \geq 0\)), and 5 (\(y \geq 0\)) are nonbinding as they do not directly affect the optimal solution.
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