To solve this linear programming problem, we must first identify the feasible region defined by the constraints and then find the value of $z = 2x + 3y$ at the corner points of the feasible region.

Problem setup:

• Objective function: Maximize (and minimize) $z = 2x + 3y$
• Subject to the constraints:
  1. $2x + y \geq 30$
  2. $x + 2y \geq 24$
  3. $x \geq 0, y \geq 0$ (non-negativity constraints)

Step 1: Solve the system of inequalities

To find the corner points, we first solve the system of inequalities by treating them as equalities.

1. Equation 1: $2x + y = 30$

   Solve for $y$ in terms of $x$: $y = 30 - 2x$

2. Equation 2: $x + 2y = 24$

   Solve for $y$ in terms of $x$: $y = \frac{24 - x}{2}$

Step 2: Find the intersection points

To find the corner points, we solve the system by finding the intersection of the two lines $2x + y = 30$ and $x + 2y = 24$.

Substitute $y = 30 - 2x$ into $x + 2y = 24$:

$x + 2(30 - 2x) = 24$ $x + 60 - 4x = 24$ $-3x + 60 = 24$ $-3x = 24 - 60 = -36$ $x = \frac{-36}{-3} = 12$

Now substitute $x = 12$ back into the equation $y = 30 - 2x$: $y = 30 - 2(12) = 30 - 24 = 6$

So, one corner point is $(12, 6)$.

Step 3: Identify boundary points

We check the boundary points by analyzing the intercepts:

• For $2x + y = 30$:

  • When $x = 0$, $y = 30$
  • When $y = 0$, $2x = 30 \Rightarrow x = 15$

• For $x + 2y = 24$:

  • When $x = 0$, $2y = 24 \Rightarrow y = 12$
  • When $y = 0$, $x = 24$

Now, let's summarize the corner points:

• Intersection: $(12, 6)$
• Intercepts: $(15, 0)$, $(0, 30)$, $(0, 12)$

Step 4: Evaluate $z = 2x + 3y$ at each corner point

1. At $(12, 6)$: $z = 2(12) + 3(6) = 24 + 18 = 42$

2. At $(15, 0)$: $z = 2(15) + 3(0) = 30$

3. At $(0, 30)$: $z = 2(0) + 3(30) = 90$

4. At $(0, 12)$: $z = 2(0) + 3(12) = 36$

Step 5: Conclusion

The maximum value of $z$ occurs at $(0, 30)$, where $z = 90$.

Thus, the correct choice is:

• A: $z = 90$

Would you like a detailed explanation of any part of the solution?

Here are 5 related questions for further practice:

1. How would the solution change if the objective function was $z = 5x + 2y$?
2. What is the minimum value of $z = 2x + 3y$ in the same region?
3. How do we handle linear programming problems with more than two variables?
4. Can you solve for the feasible region graphically for this problem?
5. What happens if one or more constraints are strict inequalities?

Tip: Always check the corner points of the feasible region in linear programming to find the optimal solution.