To solve this linear programming problem and maximize $z = 9x + 4y$, we will use the method of graphing the constraints to find the feasible region and then evaluate $z$ at the corner points.

Step 1: Convert the inequalities into equalities to find the boundary lines of the constraints.

1. Constraint 1: $2x - 5y \leq 20$

   • Rewrite as $2x - 5y = 20$.

2. Constraint 2: $-2x + y \leq 4$

   • Rewrite as $-2x + y = 4$.

3. Non-negativity constraints: $x \geq 0, y \geq 0$.

Step 2: Solve for the intercepts and plot the lines.

1. For $2x - 5y = 20$:

   • Set $x = 0$, solve for $y$: $-5y = 20$, so $y = -4$.
   • Set $y = 0$, solve for $x$: $2x = 20$, so $x = 10$.

   The intercepts are $(10, 0)$ and $(0, -4)$.

2. For $-2x + y = 4$:

   • Set $x = 0$, solve for $y$: $y = 4$.
   • Set $y = 0$, solve for $x$: $-2x = 4$, so $x = -2$.

   The intercepts are $(-2, 0)$ and $(0, 4)$.

Step 3: Find the feasible region.

Since $x \geq 0$ and $y \geq 0$, we are only interested in the first quadrant. We need to find the intersection of the lines in this region.

Step 4: Find the intersection of the two lines.

Solve the system:

1. $2x - 5y = 20$
2. $-2x + y = 4$

We can solve this by substitution or elimination. Let's solve it using substitution.

From equation (2), solve for $y$:

$y = 2x + 4$

Substitute this into equation (1):

$2x - 5(2x + 4) = 20$

$2x - 10x - 20 = 20$

$-8x - 20 = 20$

$-8x = 40 \quad \Rightarrow \quad x = -5$

Now, substitute $x = -5$ into $y = 2x + 4$:

$y = 2(-5) + 4 = -10 + 4 = -6$

Thus, the lines intersect at $(-5, -6)$, but this point is not in the first quadrant. Therefore, we will only focus on the corner points in the first quadrant.

Step 5: Evaluate $z = 9x + 4y$ at the corner points of the feasible region.

We need to evaluate $z = 9x + 4y$ at the relevant corner points.

Corner points:

1. $(0, 0)$
2. $(10, 0)$
3. Intersection of the lines in the first quadrant (to be found).

Next, let's calculate the values.The corner points are:

1. $(0, 0)$, where $z = 0$
2. $(10, 0)$, where $z = 90$

We still need to find the intersection in the first quadrant.

The lines $2x - 5y = 20$ and $-2x + y = 4$ intersect outside of the first quadrant, which means this point doesn't contribute to the feasible region. Therefore, we only consider $(0, 0)$ and $(10, 0)$.

Conclusion:

• The maximum value of $z$ is 90 at $(x, y) = (10, 0)$.

Would you like more details on how the feasible region was determined, or do you have any other questions?

Here are 5 related questions for further practice:

1. How do you graphically solve a linear programming problem?
2. What is the significance of the corner points in linear programming?
3. How can the Simplex method be applied to this problem?
4. How do changes in the objective function affect the optimal solution?
5. How can you solve a linear programming problem using matrix methods?

Tip: Always check the feasible region for boundary conditions when solving linear programming problems!