Maximize Z = 5x + 4y: Graphical Solution

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Hey guys! Today, we're diving into a classic linear programming problem and cracking it using the graphical method. It's like solving a puzzle where we need to find the best possible outcome given some constraints. Let's get started and make it super easy to understand!

Understanding the Problem

First off, let's break down what we're dealing with. We're trying to maximize the objective function Z = 5x + 4y. Think of 'Z' as the thing we want to make as big as possible – maybe it's profit, maybe it's happiness (if we could put a number on it!). The 'x' and 'y' are our decision variables, the things we can control to influence 'Z'.

Now, we can't just make 'x' and 'y' infinitely large because we have constraints. These are like rules that keep us in check. We've got two main ones:

  1. 5x + 2.5y ≤ 80
  2. x + 1.5y ≤ 70

And, of course, we can't have negative 'x' or 'y' (unless we're dealing with some seriously weird stuff), so we also have:

x ≥ 0, y ≥ 0

These constraints define a feasible region – the area where our solutions can live. Our mission is to find the point within this region that makes 'Z' the biggest it can be. Time to roll up our sleeves and get graphical!

Step-by-Step Graphical Solution

1. Plot the Constraints

Alright, the first thing we need to do is plot our constraints on a graph. This will help us visualize the feasible region. Let's take each constraint one by one.

Constraint 1: 1.5x + 2.5y ≤ 80

To plot this, we'll first treat it as an equation: 1.5x + 2.5y = 80. We need to find two points on this line. The easiest way to do that is to set x = 0 and solve for y, then set y = 0 and solve for x.

  • If x = 0: 2.5y = 80 => y = 80 / 2.5 = 32. So, one point is (0, 32).
  • If y = 0: 1.5x = 80 => x = 80 / 1.5 = 53.33 (approximately). So, another point is (53.33, 0).

Plot these two points (0, 32) and (53.33, 0) on your graph and draw a line through them. Since the constraint is ≤ 80, we want the region below this line. Shade that area.

Constraint 2: 2x + 1.5y ≤ 70

Same drill here. Turn the inequality into an equation: 2x + 1.5y = 70.

  • If x = 0: 1.5y = 70 => y = 70 / 1.5 = 46.67 (approximately). So, one point is (0, 46.67).
  • If y = 0: 2x = 70 => x = 70 / 2 = 35. So, another point is (35, 0).

Plot these two points (0, 46.67) and (35, 0) and draw a line through them. Again, since the constraint is ≤ 70, shade the region below this line.

Non-negativity Constraints: x ≥ 0, y ≥ 0

These are simple. x ≥ 0 means we're only looking at the right side of the y-axis, and y ≥ 0 means we're only looking at the upper side of the x-axis. So, we're in the first quadrant.

2. Identify the Feasible Region

The feasible region is the area where all shaded regions overlap. It's the area that satisfies all our constraints simultaneously. Look for the polygon formed by the intersection of all the shaded areas, including the first quadrant.

3. Find the Corner Points

The corner points (or vertices) of the feasible region are crucial. These are the points where the constraint lines intersect. We need to find the coordinates of all these corner points.

  • (0, 0): The origin is always a corner point when x ≥ 0 and y ≥ 0.
  • (35, 0): This is the x-intercept of the second constraint line.
  • (0, 32): This is the y-intercept of the first constraint line.
  • Intersection of the two constraint lines: To find this, we need to solve the system of equations:
    • 5x + 2.5y = 80
    • x + 1.5y = 70

Let's solve this. Multiply the first equation by 2 and the second by 1.5 to eliminate 'x':

  • 3x + 5y = 160
  • x + 2.25y = 105

Multiply the second equation by -1 and add them:

  1. 25y = 55

y = 55 / 0.25 = 22

Now, substitute y = 22 into one of the original equations, let's use 2x + 1.5y = 70:

x + 1.5(22) = 70

x + 33 = 70

x = 70 - 33 = 37

So, the intersection point is (17, 22).

4. Evaluate the Objective Function at Each Corner Point

Now, we're going to plug each of these corner points into our objective function Z = 5x + 4y and see what we get.

  • (0, 0): Z = 5(0) + 4(0) = 0
  • (35, 0): Z = 5(35) + 4(0) = 175
  • (0, 32): Z = 5(0) + 4(32) = 128
  • (17, 22): Z = 5(17) + 4(22) = 85 + 88 = 173

5. Determine the Optimal Solution

The optimal solution is the corner point that gives us the highest value of Z. Looking at our results:

  • (0, 0) gives Z = 0
  • (35, 0) gives Z = 175
  • (0, 32) gives Z = 128
  • (17, 22) gives Z = 173

The maximum value of Z is 175, which occurs at the point (35, 0).

Conclusion

So, to maximize Z = 5x + 4y subject to our constraints, we should set x = 35 and y = 0. This gives us a maximum Z value of 175. And that's how you solve a linear programming problem using the graphical method! It's all about plotting the constraints, finding the feasible region, identifying the corner points, and then testing those points in the objective function. Easy peasy, right?

Remember, this method is super helpful for visualizing what's going on. For more complex problems with more variables, you might need to use other methods like the simplex algorithm, but the graphical method is a fantastic way to get your head around the basics. Keep practicing, and you'll be a pro in no time!