Question

The minimum of the following linear programming problem occurs at:
Minimize $C = 7x + 10y$
subject to $x + y \geq 3$, $x + 2y \geq 4$, $x, y \geq 0$

• A. $(3,0)$
• B. $(4,0)$
• C. $(0,2)$
• D. $(2,1)$
• E. $(0,3)$

Hint:

For minimization problems with "greater than or equal to" ($\geq$) constraints, the optimal point is usually the intersection of the constraint lines. Evaluate this point first!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The optimal solution (minimum or maximum) of a linear programming problem occurs at one of the corner points of the feasible region defined by the constraints.

Step 2: Detailed Explanation:
1. Determine the corner points of the feasible region:
- Intersection of $x+y=3$ and $x+2y=4$:
Subtracting the first from the second gives $y=1$. Then $x=2$. Point: $(2, 1)$.
- $x$-intercepts of constraints: $x=3$ and $x=4$. Since constraints are $\geq$, the larger intercept $x=4$ is on the boundary. Point: $(4, 0)$.
- $y$-intercepts of constraints: $y=3$ and $y=2$. Since constraints are $\geq$, the larger intercept $y=3$ is on the boundary. Point: $(0, 3)$.
2. The feasible region is unbounded and has corner points: $(4,0), (2,1), (0,3)$.
3. Evaluate the objective function $C = 7x + 10y$ at these corner points:
- At $(4, 0)$: $C = 7(4) + 10(0) = 28$.
- At $(2, 1)$: $C = 7(2) + 10(1) = 24$.
- At $(0, 3)$: $C = 7(0) + 10(3) = 30$.
Comparing the values, the minimum value is 24, occurring at $(2, 1)$.

Step 3: Final Answer:
The minimum occurs at the point $(2,1)$.