To solve the problem of maximizing the linear function [tex]\( z = 5x + 3y \)[/tex] subject to the given constraints, follow these steps:
### Constraints
1. [tex]\( 2x + 7y \leq 42 \)[/tex]
2. [tex]\( 4x + 3y \leq 40 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Steps to find the solution:
#### 1. Identify the feasible region
The feasible region is determined by the intersection of the half-planes defined by the constraints. We need to find the boundary lines of these constraints:
- For [tex]\( 2x + 7y = 42 \)[/tex]:
- When [tex]\( x = 0 \)[/tex]: [tex]\( 7y = 42 \implies y = 6 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( 2x = 42 \implies x = 21 \)[/tex]
So, the line [tex]\( 2x + 7y = 42 \)[/tex] intersects the axes at (21, 0) and (0, 6).
- For [tex]\( 4x + 3y = 40 \)[/tex]:
- When [tex]\( x = 0 \)[/tex]: [tex]\( 3y = 40 \implies y = \frac{40}{3} \approx 13.\overline{3} \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( 4x = 40 \implies x = 10 \)[/tex]
So, the line [tex]\( 4x + 3y = 40 \)[/tex] intersects the axes at (10, 0) and (0, 13.\overline{3}).
#### 2. Determine the vertices of the feasible region
The vertices of the feasible region are determined by the intersection points of the boundary lines:
We have the points of intersection:
- (0, 0)
- (21, 0)
- (0, 6)
- (10, 0)
- The intersection of [tex]\( 2x + 7y = 42 \)[/tex] and [tex]\( 4x + 3y = 40 \)[/tex].
To find the intersection of [tex]\( 2x + 7y = 42 \)[/tex] and [tex]\( 4x + 3y = 40 \)[/tex], we solve:
[tex]\[
\begin{cases}
2x + 7y = 42 \\
4x + 3y = 40
\end{cases}
\][/tex]
Multiply the first equation by 2:
[tex]\[
4x + 14y = 84
\][/tex]
Subtract the second equation:
[tex]\[
4x + 14y - (4x + 3y) = 84 - 40 \\
11y = 44 \\
y = 4
\][/tex]
Substituting [tex]\( y = 4 \)[/tex] into the first equation:
[tex]\[
2x + 7(4) = 42 \\
2x + 28 = 42 \\
2x = 14 \\
x = 7
\][/tex]
So, the intersection point is (7, 4).
#### 3. Evaluate the objective function at each vertex
We evaluate [tex]\( z = 5x + 3y \)[/tex] at each vertex:
- At (0, 0): [tex]\( z = 5(0) + 3(0) = 0 \)[/tex]
- At (21, 0): [tex]\( z = 5(21) + 3(0) = 105 \)[/tex]
- At (0, 6): [tex]\( z = 5(0) + 3(6) = 18 \)[/tex]
- At (10, 0): [tex]\( z = 5(10) + 3(0) = 50 \)[/tex]
- At (7, 4): [tex]\( z = 5(7) + 3(4) = 35 + 12 = 47 \)[/tex]
#### Conclusion
The maximum value of [tex]\( z \)[/tex] occurs at [tex]\( (10, 0) \)[/tex] and it is [tex]\( 50 \)[/tex].
Thus, the solution is:
Maximum is [tex]\( \boxed{50} \)[/tex] at
[tex]\[
\begin{array}{l}
x = \boxed{10} \\
y = \boxed{0}
\end{array}
\][/tex]
