Example Question - linear programming
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Optimizing Linear Programming with Constraints
The image contains a linear programming problem that asks to minimize and maximize the objective function P = 25x + 6y given the constraints: 1. x + y ≤ 23 2. 4x + 5y ≤ 43 3. x ≥ 5 4. y ≥ 0 Firstly, we want to find the feasible region defined by the constraints and then locate the corner points since the minimum and maximum values for P will occur at the corners of this region due to the nature of linear programming problems. Let's find the intersection points of the lines given by the constraints: From the first constraint, if x = 5 (the minimum value for x), we can find the maximum y: 5 + y ≤ 23 → y = 18 From the second constraint, if x = 5, we can find y: 4(5) + 5y ≤ 43 → 20 + 5y ≤ 43 → 5y ≤ 23 → y = 4.6 (but since y needs to be an integer, y = 4) Now let's see where the first and second constraints intersect: x + y = 23 and 4x + 5y = 43 can be solved as a system of equations. However, rather than solving it algebraically, it might be easier to graph these constraints and visually find the intersection points, especially since we know they have to be integers due to the contexts of these problems. Graphing these lines, along with x ≥ 5 and y ≥ 0, we notice that the possible intersection points (i.e., the corner points of the feasible region) are: - (5, 18) from x = 5 and x + y = 23 - (5, 4) from x = 5 and 4x + 5y = 43 - The intersection of x + y = 23 and 4x + 5y = 43, which we will calculate. Solving the system of equations: x + y = 23 4x + 5y = 43 Multiply the first equation by 4: 4x + 4y = 92 4x + 5y = 43 Subtract the second equation from the multiplied first one: 4y - 5y = 92 - 43 -y = 49 y = -49, which is not possible since y ≥ 0, so this intersection is not within the feasible region. Thus, the corner points of the feasible region are (5, 18) and (5, 4) because the intersection of the two constraints falls outside the feasible region. Let's now find the values of P at these corner points: For (5, 18): P = 25(5) + 6(18) = 125 + 108 = 233 For (5, 4): P = 25(5) + 6(4) = 125 + 24 = 149 Now, comparing the values, we see that the minimum value of P is 149 at the point (5, 4). Answering the questions: - What is the minimum value of P? A: 149 - What are the coordinates of the corner point where the minimum value of P occurs? A: (5, 4)
Linear Programming: Maximizing Objective Function
This problem is asking for the maximum value of the objective function \( 2x + y \) within the constraints described by the inequalities and the shaded feasible region on the graph. The constraints are: 1. \( y \geq 2x \) (Above the line \( y = 2x \)) 2. \( y \geq x \) (Above the line \( y = x \)) 3. \( x + y \leq 6 \) (Below the line \( x + y = 6 \)) The shaded region on the graph represents the set of all points (x, y) that satisfy these constraints. To find the maximum value of the objective function, we need to evaluate it at each of the vertices of the feasible region, as the maximum value in a linear programming problem always occurs at a vertex of the feasible region. From the graph, it is clear that there are three vertices of the shaded feasible region: 1. \( P \) at the intersection of \( y = x \) and \( x + y = 6 \). 2. \( Q \) at (3, 3), which is provided in the problem statement. 3. \( R \) at the intersection of \( y = 2x \) and \( x + y = 6 \). First, we need to find the coordinates of the vertex \( P \) and \( R \): - For \( P \): When \( y = x \), we substitute \( y \) for \( x \) in \( x + y = 6 \) to get \( x + x = 6 \) or \( 2x = 6 \), so that \( x = 3 \). Thus, \( P \) is at (3, 3). The value of \( 2x + y \) at \( P \) is \( 2(3) + 3 = 9 \). - For \( R \): To find \( R \), solve the system of equations \( y = 2x \) and \( x + y = 6 \) simultaneously. Replacing \( y \) in the second equation with \( 2x \) gives \( x + 2x = 6 \), which simplifies to \( 3x = 6 \), so \( x = 2 \). Substituting back into \( y = 2x \), we find \( y = 4 \); thus, \( R \) is at (2, 4). The value of \( 2x + y \) at \( R \) is \( 2(2) + 4 = 8 \). We were given that \( Q \) is at (3, 3), which is the same point as \( P \), and we have already calculated the objective function value at \( P \), which is 9. Now, compare the values of the objective function at these vertices: - At \( P \) and \( Q \) (which are the same point), the value of \( 2x + y \) is 9. - At \( R \), the value of \( 2x + y \) is 8. The maximum value is therefore 9. The correct answer to the problem is: A: 9
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