Question

Consider the Linear Programming Problem, where the objective function \[ Z = x + 4y \] needs to be minimized subject to the following constraints: \[ 2x + y \geq 1000, \] \[ x + 2y \geq 800, \] \[ x \geq 0, \quad y \geq 0. \] Draw a neat graph of the feasible region and find the minimum value of $Z$.

Hint:

To find the minimum or maximum of the objective function in a linear programming problem, evaluate the objective function at each corner point of the feasible region and choose the one that gives the required extreme value.

Answer 1

We are asked to minimize the objective function $Z = x + 4y$ subject to the given constraints. First, let’s write down the constraints and find the feasible region. 1. Plot the constraints: - From $2x + y \geq 1000$, we get the line $y = 1000 - 2x$. This represents the constraint $2x + y \geq 1000$. - From $x + 2y \geq 800$, we get the line $y = \frac{800 - x}{2}$. This represents the constraint $x + 2y \geq 800$. - The constraints $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant. 2. Find the intersection points of the lines: We solve the system of equations given by the two constraints to find the intersection points. \[ \text{Equation 1: } 2x + y = 1000 \] \[ \text{Equation 2: } x + 2y = 800 \] Multiply Equation 2 by 2 to make the coefficient of $x$ equal: \[ 2x + 4y = 1600 \] Now subtract Equation 1 from this equation: \[ (2x + 4y) - (2x + y) = 1600 - 1000 \] \[ 3y = 600 \quad \Rightarrow \quad y = 200. \] Substitute $y = 200$ into Equation 1: \[ 2x + 200 = 1000 \quad \Rightarrow \quad 2x = 800 \quad \Rightarrow \quad x = 400. \] Thus, the point of intersection is $(400, 200)$. 3. Check the boundary points: The feasible region is bounded by the x-axis ($y = 0$) and the y-axis ($x = 0$). We now check the intersection of each constraint with the axes. - When $x = 0$ in Equation 1: \[ 2(0) + y = 1000 \quad \Rightarrow \quad y = 1000. \] So, the point is $(0, 1000)$. - When $y = 0$ in Equation 2: \[ x + 2(0) = 800 \quad \Rightarrow \quad x = 800. \] So, the point is $(800, 0)$. 4. Graph the feasible region: Plot the lines for $2x + y = 1000$ and $x + 2y = 800$ on the coordinate plane. The feasible region is the area bounded by these lines, the x-axis, and the y-axis. 5. Objective function at the corner points: The corner points of the feasible region are: - $(0, 1000)$ - $(400, 200)$ - $(800, 0)$ We now substitute these points into the objective function $Z = x + 4y$. - At $(0, 1000)$: \[ Z = 0 + 4(1000) = 4000. \] - At $(400, 200)$: \[ Z = 400 + 4(200) = 400 + 800 = 1200. \] - At $(800, 0)$: \[ Z = 800 + 4(0) = 800. \] The minimum value of $Z$ is $800$ at the point $(800, 0)$. 6. Conclusion: The minimum value of the objective function $Z = x + 4y$ is $800$, and this occurs at the point $(800, 0)$.