Example Question - optimization
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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)
Optimizing Linear Programming Problem
The question in the image shows a linear programming problem where the objective is to maximize the function Z = 10X + 12Y subject to three constraints: 1. 2X + 3Y ≤ 1500 2. 3X + 2Y ≤ 1500 3. X + Y ≤ 600 and the non-negativity constraints: X ≥ 0 Y ≥ 0 To solve this optimization problem, follow these steps: 1. **Graph the Constraints:** Plot the lines represented by each constraint on a graph. The area that satisfies all constraints is known as the feasible region. 2. **Find the Intersection Points of the Feasible Region:** Calculate the intersection points of the lines forming the boundaries of the feasible region. You'll need to solve the equations where two lines intersect. This includes the intersections with the X and Y axes. 3. **Evaluate the Objective Function at each Intersection Point:** That is, substitute the values of X and Y obtained from the intersection points into the objective function Z = 10X + 12Y to find out which point gives the maximum value of Z. Let's calculate the intersection points: For the X-axis (Y=0), using constraint 3 (X + Y ≤ 600): X = 600, Y = 0 For the Y-axis (X=0), using constraint 3 (X + Y ≤ 600): Y = 600, X = 0 Now let's find intersection of the lines by pairing the constraints: For the first two constraints: 2X + 3Y = 1500 3X + 2Y = 1500 You can solve these equations simultaneously or use methods like substitution or elimination to find the values of X and Y. Let's use the elimination method to solve it straightforwardly: Multiply the first equation by 3 and the second equation by 2: 6X + 9Y = 4500 6X + 4Y = 3000 Now, subtract the second equation from the first: 6X + 9Y - (6X + 4Y) = 4500 - 3000 5Y = 1500 Y = 300 Substitute Y = 300 into the first original constraint: 2X + 3(300) ≤ 1500 2X + 900 ≤ 1500 2X ≤ 600 X = 300 So we have another intersection point at (300, 300). Now you will need to find the intersection between the last two constraints 3X + 2Y = 1500 and X + Y = 600 in a similar manner, and then you will need to calculate the values of the objective function Z at each of these intersection points, including the intersections with the axes: (600, 0) and (0, 600). The maximum value of Z will determine the optimal solution for the problem. If that maximum occurs at any of the corners of the feasible region, then that point represents the maximum values of X and Y for the problem.
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