91Ó°ÊÓ

We need to find the corner points of the feasible region. First, let's rewrite the inequalities as equations: $$ x+2y = 4 \\ x+y = 3 \\ x = 0, y = 0 $$

Now we'll find the intersection points of the lines derived from the inequalities, and these will be our corner points of the feasible region. We have four possible intersections: (1) Between the first and second lines, (2) between the first line and x-axis, (3) between the second line and x-axis, and (4) between the second line and y-axis. Intersection of the first and second lines: $$ \begin{cases} x + 2y = 4 \\ x + y = 3 \\ \end{cases} $$ Solving these equations, we get the point (1, 2). Intersection of the first line and x-axis: $$ \begin{cases} x + 2y = 4 \\ x = 0 \\ \end{cases} $$ Solving these equations, we get the point (0, 2). Intersection of the second line and x-axis: $$ \begin{cases} x + y = 3 \\ x = 0 \\ \end{cases} $$ Solving these equations, we get the point (0, 3). Intersection of the second line and y-axis: $$ \begin{cases} x + y = 3 \\ y = 0 \\ \end{cases} $$ Solving these equations, we get the point (3, 0).

Now, we will substitute these corner points into our objective function \(C = 2x + 5y\) and find the minimum value: $$ C(1, 2) = 2(1) + 5(2) = 12 \\ C(0, 2) = 2(0) + 5(2) = 10 \\ C(0, 3) = 2(0) + 5(3) = 15 \\ C(3, 0) = 2(3) + 5(0) = 6 $$ The minimum value of C occurs at the point (3, 0) with the value of 6. a. The optimal solution using the method of corners is (3, 0) with a minimum cost of \(C = 6\). We will now address the other parts of the problem: b. The optimal solution occurs at the intersection of the second line (\(x+y \geq 3\)) and the x-axis. The range of values for the coefficient of \(x\) in the objective function, such that it doesn't change the optimal solution, is when the second constraint remains binding. To keep it binding, we must ensure that changing the coefficient of \(x\) does not make the first constraint become the new binding constraint. Therefore, we must ensure that the slope of the objective function, \(-\frac{2}{5}\), remains between the slopes of the first two constraints, which are \(-\frac{1}{2}\) and \(-1\): $$ -1 < -\frac{2}{a} < -\frac{1}{2} $$ Solving for \(a\), we get \(\frac{4}{3} < a < 2\). b. The range of values that the coefficient of \(x\) can assume without changing the optimal solution is \(\frac{4}{3} < a < 2\). c. Resource 1 (requirement 1) corresponds to the first constraint, \(x + 2y \geq 4\). To find the range of values it can assume without affecting the optimal solution, we'll consider the constraint as \(x + 2y \geq k\) and observe that the intersection of this constraint and the second constraint remains at the optimal point (3, 0): $$ (3) + 2(0) \geq k $$ k can be as low as 3, but there is no upper bound. c. The range of values that resource 1 (requirement 1) can assume is \(k \geq 3\). d. The shadow price for resource 1 (requirement 1) is the change in the objective function when the constraint is increased by one unit. Since the range for resource 1 showed no upper bound, resource 1 doesn't affect the optimal solution, and its shadow price is 0. d. The shadow price for resource 1 (requirement 1) is 0. e. To identify binding and nonbinding constraints, we can observe how the constraints affect the optimal solution. A constraint is binding if it affects the optimal solution. The second constraint (\(x+y\geq3\)) is binding, as it determines the optimal solution. The first constraint (\(x+2y\geq4\)) is nonbinding, as it doesn't affect the optimal solution. e. Binding constraint: \(x+y\geq3\), Nonbinding constraint: \(x+2y\geq4\)