Solve graphically

max z=2x+1y

when 2x+1y≤2

4x+3y≤5
x,y≥0

Give an exact answer or round it to one decimal place.

To solve this linear programming problem graphically, we need to follow these steps:

### 1. **Graph the constraints:**
   We will first convert the inequalities into equalities, then plot the lines, and finally identify the feasible region. The constraints are:

   \[
   2x + y \leq 2
   \]
   \[
   4x + 3y \leq 5
   \]
   \[
   x \geq 0, \quad y \geq 0
   \]

### 2. **Find the intersection points of the lines:**

#### Constraint 1: \( 2x + y = 2 \)

   - When \( x = 0 \), \( y = 2 \).
   - When \( y = 0 \), \( 2x = 2 \), so \( x = 1 \).

   Therefore, the line passes through points \( (0, 2) \) and \( (1, 0) \).

#### Constraint 2: \( 4x + 3y = 5 \)

   - When \( x = 0 \), \( 3y = 5 \), so \( y = \frac{5}{3} \approx 1.67 \).
   - When \( y = 0 \), \( 4x = 5 \), so \( x = \frac{5}{4} = 1.25 \).

   Therefore, the line passes through points \( (0, 1.67) \) and \( (1.25, 0) \).

### 3. **Determine the feasible region:**
   The feasible region is the area that satisfies all the inequalities. This is the region where the two shaded areas from the inequalities overlap, and it is bounded by the lines we just plotted, along with the positive \( x \)- and \( y \)-axes.

### 4. **Find the corner points:**
   The optimal solution will be at one of the corner points of the feasible region, so we need to calculate the intersection of the two constraint lines.

   Solve the system of equations:
   \[
   2x + y = 2
   \]
   \[
   4x + 3y = 5
   \]

   - Multiply the first equation by 3 to eliminate \( y \):
     \[
     6x + 3y = 6
     \]
   - Subtract the second equation from this:
     \[
     (6x + 3y) - (4x + 3y) = 6 - 5
     \]
     \[
     2x = 1 \implies x = \frac{1}{2} = 0.5
     \]
   - Substitute \( x = 0.5 \) into \( 2x + y = 2 \):
     \[
     2(0.5) + y = 2 \implies 1 + y = 2 \implies y = 1
     \]

   Thus, the lines intersect at \( (0.5, 1) \).

### 5. **Evaluate the objective function at the corner points:**

   The corner points of the feasible region are \( (0, 0) \), \( (1, 0) \), \( (0, 1.67) \), and \( (0.5, 1) \). We will now evaluate \( z = 2x + y \) at each of these points:

   - At \( (0, 0) \): \( z = 2(0) + 0 = 0 \)
   - At \( (1, 0) \): \( z = 2(1) + 0 = 2 \)
   - At \( (0, 1.67) \): \( z = 2(0) + 1.67 = 1.67 \)
   - At \( (0.5, 1) \): \( z = 2(0.5) + 1 = 2 \)

### 6. **Conclusion:**
   The maximum value of \( z \) occurs at both \( (1, 0) \) and \( (0.5, 1) \), and the maximum value is \( \mathbf{z = 2} \).

### Final Answer:
The maximum value of \( z \) is \( 2 \), and it occurs at the points \( (1, 0) \) and \( (0.5, 1) \).

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Would you like further details on any step?

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### Related Questions:
1. How do you graph inequalities on a coordinate plane?
2. What is the significance of corner points in linear programming?
3. How can we solve linear programming problems using the Simplex method?
4. How would the solution change if one of the constraints is removed?
5. What happens if the objective function coefficients are changed?

**Tip:** When solving linear programming problems graphically, always check all corner points of the feasible region since the optimal solution will always be at one of them.

Solve graphically max z=2x+1y when 2x+1y≤2, 4x+3y≤5, x,y≥0. Give an exact answer or round it to one decimal place.

This guide explains how to solve the linear programming problem of maximizing z = 2x + y, given the constraints 2x + y ≤ 2, 4x + 3y ≤ 5, and x, y ≥ 0. Learn how to graph the inequalities, find the feasible region, and calculate the maximum value at corner points. Suitable for students in Grades 11-12.

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