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Problem 1

# 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}

Expert verified
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 $$C = -12$$.
See the step by step solution

## 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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