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Use the technique developed in this section to solve the minimization problem. $$ \begin{aligned} \text { Minimize } & C=-2 x+y \\ \text { subject to } & x+2 y \leq 6 \\ & 3 x+2 y \leq 12 \\ & x \geq 0, y \geq 0 \end{aligned} $$

Short Answer

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The feasible region for the given constraints is a quadrilateral with corner points \((0, 3)\), \((6, 0)\), \((4, 0)\), and \((3, 1.5)\). Evaluating the objective function \(C = -2x + y\) at each corner point, we find the minimum value \(C = -12\) at the point \((6, 0)\). Therefore, the solution to the minimization problem is \(x = 6\) and \(y = 0\) with a minimum value of \(C = -12\).
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Step 1: Plot the feasible region

To find the feasible region, we will plot the inequalities on a coordinate plane: 1. \(x+2 y \leq 6\) 2. \(3 x+2 y \leq 12\) 3. \(x \geq 0\) 4. \(y \geq 0\) First, turn the inequalities into equalities and solve for \(y\): 1. \(y = -\frac{1}{2}x + 3\) 2. \(y = -\frac{3}{2}x + 6\) 3. The vertical line \(x=0\) (y-axis). 4. The horizontal line \(y=0\) (x-axis). Now plot these lines on a graph. The area where all the inequalities are satisfied is where the feasible region lies.

Step 2: Identify the corner points of the feasible region

The feasible region is a quadrilateral with vertices at the following points: 1. The intersection between lines 1 and 3: \((0, 3)\) 2. The intersection between lines 1 and 4: \((6, 0)\) 3. The intersection between lines 2 and 4: \((4, 0)\) 4. The intersection between lines 1 and 2: Solve the system of equations: \(\begin{cases} y = -\frac{1}{2}x + 3 \\ y = -\frac{3}{2}x + 6 \\ \end{cases}\) Solving this system, we find that \(x=3\) and \(y=1.5\). So, the fourth corner point is \((3,1.5)\).

Step 3: Evaluate the objective function at each corner point

Now we have to find the value of the objective function \(C=-2x+y\) at each corner point: 1. \(C(0, 3) = -2(0) + 3 = 3\) 2. \(C(6, 0) = -2(6) + 0 = -12\) 3. \(C(4, 0) = -2(4) + 0 = -8\) 4. \(C(3, 1.5) = -2(3) +1.5= -4.5\)

Step 4: Determine the minimum value of the objective function

Among the values obtained in step 3, the minimum value is \(C = -12\), which occurs at the point \((6, 0)\). So, the solution to the minimization problem is \(x = 6\) and \(y = 0\) with the minimum value of \(C = -12\).

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Most popular questions from this chapter

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