To solve the given linear programming problem, we need to maximize the objective function [tex]\( z = 3x + 4y \)[/tex] subject to the constraints:
[tex]\[
\begin{align*}
1. & \quad 2x + 6y \leq 54 \\
2. & \quad 7x + y \leq 49 \\
3. & \quad x \geq 0, y \geq 0
\end{align*}
\][/tex]
Here's a step-by-step breakdown of how to find the maximum value of [tex]\( z \)[/tex] and the corresponding values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
1. Graph the constraints:
- Convert the inequalities into equalities to find the boundary lines.
- For [tex]\( 2x + 6y = 54 \)[/tex]:
- When [tex]\( x = 0 \)[/tex]: [tex]\( 6y = 54 \)[/tex] ⟹ [tex]\( y = 9 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( 2x = 54 \)[/tex] ⟹ [tex]\( x = 27 \)[/tex]
- For [tex]\( 7x + y = 49 \)[/tex]:
- When [tex]\( x = 0 \)[/tex]: [tex]\( y = 49 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( 7x = 49 \)[/tex] ⟹ [tex]\( x = 7 \)[/tex]
2. Find the feasible region:
- The feasible region lies where the constraints overlap in the first quadrant (since [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex]).
3. Identify the corner points:
- Solve the system of equations formed by the intersection points:
- Intersection of [tex]\( 2x + 6y = 54 \)[/tex] and [tex]\( 7x + y = 49 \)[/tex]:
- Multiply the second equation by 6: [tex]\( 42x + 6y = 294 \)[/tex]
- Subtract the first equation from this: [tex]\( 42x + 6y - 2x - 6y = 294 - 54 \)[/tex] ⟹ [tex]\( 40x = 240 \)[/tex] ⟹ [tex]\( x = 6 \)[/tex]
- Substitute [tex]\( x = 6 \)[/tex] into [tex]\( 7x + y = 49 \)[/tex]: [tex]\( 7(6) + y = 49 \)[/tex] ⟹ [tex]\( 42 + y = 49 \)[/tex] ⟹ [tex]\( y = 7 \)[/tex]
- So, their intersection point is [tex]\( (6, 7) \)[/tex].
- Other corner points are obtained from setting [tex]\( x \)[/tex] or [tex]\( y \)[/tex] to 0:
- [tex]\( (0, 0) \)[/tex]
- [tex]\( (0, 9) \)[/tex]
- [tex]\( (7, 0) \)[/tex]
4. Evaluate the objective function at each corner point:
- [tex]\( z = 3x + 4y \)[/tex]
- At [tex]\( (0, 0) \)[/tex]: [tex]\( z = 3(0) + 4(0) = 0 \)[/tex]
- At [tex]\( (0, 9) \)[/tex]: [tex]\( z = 3(0) + 4(9) = 36 \)[/tex]
- At [tex]\( (7, 0) \)[/tex]: [tex]\( z = 3(7) + 4(0) = 21 \)[/tex]
- At [tex]\( (6, 7) \)[/tex]: [tex]\( z = 3(6) + 4(7) = 18 + 28 = 46 \)[/tex]
5. Determine the maximum value:
- The maximum value of [tex]\( z \)[/tex] is found at the point [tex]\( (6, 7) \)[/tex] and it is [tex]\( 46 \)[/tex].
Thus, the maximum value of [tex]\( z \)[/tex] is [tex]\( \boxed{46} \)[/tex], achieved at [tex]\( x = 6 \)[/tex] and [tex]\( y = 7 \)[/tex].
This completes the solution. If you have any further questions or need clarification on any step, feel free to ask!
