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:
This is a standard linear programming problem. The minimum value of the objective function \(C\) will occur at one of the corner points of the feasible region defined by the constraints.
Step 2: Key Formula or Approach:
Identify the vertices of the unbounded feasible region by finding the intersections of the constraint boundary lines and axes.
Evaluate \(C = 7x + 10y\) at all valid vertices.
Step 3: Detailed Explanation:
The boundary lines are:
1) \(x + y = 3\)
2) \(x + 2y = 4\)
Find their intersection:
From (1), \(x = 3 - y\). Substitute into (2):
\[ (3 - y) + 2y = 4 \implies 3 + y = 4 \implies y = 1 \] \[ x = 3 - 1 = 2 \] Intersection vertex: \((2, 1)\).
Now, find the intercepts of the constraints that enclose the feasible region (away from the origin since it's \(\ge\)).
For x-axis intercepts (\(y = 0\)):
Line 1 gives \(x = 3\). But \(3 + 2(0) = 3 \not\ge 4\), so (3,0) is not in the region.
Line 2 gives \(x = 4\). \(4 + 0 = 4 \ge 3\), so \((4,0)\) is valid.
For y-axis intercepts (\(x = 0\)):
Line 2 gives \(y = 2\). But \(0 + 2 = 2 \not\ge 3\), so (0,2) is not in the region.
Line 1 gives \(y = 3\). \(0 + 2(3) = 6 \ge 4\), so \((0,3)\) is valid.
The valid corner points are \((4,0)\), \((2,1)\), and \((0,3)\).
Evaluate \(C = 7x + 10y\) at each corner:
At \((4,0)\): \(C = 7(4) + 10(0) = 28\)
At \((2,1)\): \(C = 7(2) + 10(1) = 14 + 10 = 24\)
At \((0,3)\): \(C = 7(0) + 10(3) = 30\)
The minimum value is 24, which occurs at \((2,1)\).
Step 4: Final Answer:
The minimum occurs at (2,1).