- Determine the axis intercepts for each constraint line: \(L_1: (0, 4), (3, 0)\) and \(L_2: (0, 2.5), (7.5, 0)\).
- Solve the system of equations for the constraint lines to find their intersection at \((1.5, 2)\).
- Evaluate the constraints to bound the region: the feasible vertices are \((0, 0), (0, 2.5), (1.5, 2), (3, 0)\).
- The correct option is \(\boxed{(0, 0), (0, 2.5), (1.5, 2), (3, 0)}\).
Explanation
- Analysis of the Feasible Region
The problem asks for the vertices of the feasible region defined by the following system of linear inequalities:
\[ \begin{array}{l} 4x + 3y \leq 12 \ 2x + 6y \leq 15 \ x \geq 0 \ y \geq 0 \end{array} \]
The non-negativity constraints \(x \geq 0\) and \(y \geq 0\) indicate that the region lies entirely in the first quadrant. The boundary of this region is formed by the coordinate axes and the lines \(L_1: 4x + 3y = 12\) and \(L_2: 2x + 6y = 15\). The vertices are found by determining the intersections of these boundary lines with each other and with the coordinate axes, while ensuring the points satisfy all original inequalities.
- Determining Axis Intercepts
First, we identify the intercepts of the boundary lines:
For \(L_1: 4x + 3y = 12\):
- Setting \(x = 0\), we get \(3y = 12\), so \(y = 4\). The intercept is \((0, 4)\).
- Setting \(y = 0\), we get \(4x = 12\), so \(x = 3\). The intercept is \((3, 0)\).
For \(L_2: 2x + 6y = 15\):
- Setting \(x = 0\), we get \(6y = 15\), so \(y = 2.5\). The intercept is \((0, 2.5)\).
- Setting \(y = 0\), we get \(2x = 15\), so \(x = 7.5\). The intercept is \((7.5, 0)\).
- Finding the Intersection Point
Next, we find the intersection point of the lines \(L_1\) and \(L_2\) by solving the system:
\[ \begin{cases} 4x + 3y = 12 \ 2x + 6y = 15 \end{cases} \]
Multiply the first equation by \(2\) to align the coefficients of \(y\):
\(8x + 6y = 24\)
Subtract the second equation from this:
\((8x + 6y) - (2x + 6y) = 24 - 15\)
\(6x = 9 \implies x = 1.5\)
Substitute \(x = 1.5\) into the first equation:
\(4(1.5) + 3y = 12 \implies 6 + 3y = 12 \implies 3y = 6 \implies y = 2\)
The intersection point is \((1.5, 2)\).
- Identifying the Vertices
Finally, we determine the vertices that bound the region satisfying all four inequalities. The feasible region is bounded by the y-axis, the x-axis, and the constraint lines.
- The y-axis segment is bounded from \(y=0\) up to the smaller y-intercept of the two constraint lines, which is \((0, 2.5)\) because \((0, 2.5)\) satisfies \(4(0) + 3(2.5) = 7.5 \leq 12\).
- The x-axis segment is bounded from \(x=0\) to the smaller x-intercept, which is \((3, 0)\) because \((3, 0)\) satisfies \(2(3) + 6(0) = 6 \leq 15\).
- The intersection point \((1.5, 2)\) is a vertex as it satisfies both constraints.
- The origin \((0, 0)\) is always a vertex for these types of linear programs.
Thus, the vertices are \((0, 0)\), \((0, 2.5)\), \((1.5, 2)\), and \((3, 0)\). This corresponds to the first option.
Examples
A factory produces two types of products using a limited amount of raw materials. Product A requires 4 units of material 1 and 2 units of material 2. Product B requires 3 units of material 1 and 6 units of material 2. If the factory has a maximum of 12 units of material 1 and 15 units of material 2, the production possibilities form a feasible region. Finding the vertices of this region helps the factory determine the most efficient production mix to maximize profits within its resource constraints.