To find the maximum value of \( z = 6xy + y^2 \) under the constraints \( 4x + 3y \leq 75 \) and \( 3x + 4y \leq 100 \) with \( x, y \geq 0 \), we will follow these steps:
### Step 1: Identify the Constraints
We have two inequalities:
1. \( 4x + 3y \leq 75 \)
2. \( 3x + 4y \leq 100 \)
### Step 2: Find the Intercepts of the Lines
To graph the inequalities, we first find the intercepts of each line.
**For the first line \( 4x + 3y = 75 \):**
- When \( x = 0 \):
\[
3y = 75 \implies y = 25 \quad \text{(Point: (0, 25))}
\]
- When \( y = 0 \):
\[
4x = 75 \implies x = \frac{75}{4} = 18.75 \quad \text{(Point: (18.75, 0))}
\]
**For the second line \( 3x + 4y = 100 \):**
- When \( x = 0 \):
\[
4y = 100 \implies y = 25 \quad \text{(Point: (0, 25))}
\]
- When \( y = 0 \):
\[
3x = 100 \implies x = \frac{100}{3} \approx 33.33 \quad \text{(Point: (33.33, 0))}
\]
### Step 3: Graph the Lines
Plot the points (0, 25), (18.75, 0), and (33.33, 0) on the graph. The lines will divide the first quadrant into regions.
### Step 4: Determine the Feasible Region
Since both inequalities are satisfied at the origin (0,0), the feasible region will be in the first quadrant bounded by the lines.
### Step 5: Find the Vertices of the Feasible Region
The vertices of the feasible region can be found by solving the equations of the lines:
1. \( 4x + 3y = 75 \)
2. \( 3x + 4y = 100 \)
To find the intersection point, we can solve these equations simultaneously.
From \( 4x + 3y = 75 \):
\[
y = \frac{75 - 4x}{3}
\]
Substituting into the second equation:
\[
3x + 4\left(\frac{75 - 4x}{3}\right) = 100
\]
Multiplying through by 3 to eliminate the fraction:
\[
9x + 4(75 - 4x) = 300
\]
\[
9x + 300 - 16x = 300
\]
\[
-7x = 0 \implies x = 0
\]
Substituting \( x = 0 \) back into \( 4x + 3y = 75 \):
\[
3y = 75 \implies y = 25
\]
So one vertex is \( (0, 25) \).
### Step 6: Evaluate \( z \) at the Vertices
Now we will evaluate \( z = 6xy + y^2 \) at the vertices:
1. At \( (0, 25) \):
\[
z = 6(0)(25) + (25)^2 = 625
\]
2. At \( (18.75, 0) \):
\[
z = 6(18.75)(0) + (0)^2 = 0
\]
3. At \( (33.33, 0) \):
\[
z = 6(33.33)(0) + (0)^2 = 0
\]
### Step 7: Determine the Maximum Value
The maximum value of \( z \) occurs at the vertex \( (0, 25) \):
\[
\text{Maximum value of } z = 625
\]
### Conclusion
The maximum value of \( z \) is \( 625 \) at the point \( (0, 25) \).
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