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Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. \(\vee\) Minimize \(\quad c=3 x-3 y\) subject to \(\begin{aligned} \frac{x}{4} & \leq y \\ y & \leq \frac{2 x}{3} \\ x+y & \geq 5 \\ x+2 y & \leq 10 \\ x \geq 0, y & \geq 0 \end{aligned}\)

Short Answer

Expert verified
The feasible region for the given linear programming problem is a quadrilateral formed by the intersection of the lines \(y = \frac{x}{4}\), \(y=\frac{2x}{3}\), \(y = -x + 5\), and \(y = \frac{1}{2}(10-x)\). The corner points are (2, 0.5), (2, 3), (3, 2), and (0, 0). Evaluating the objective function c = 3x - 3y at these corner points, we find that the minimum value is -3 at the point (2, 3). So, the solution is c = -3 at the point (2, 3).

Step by step solution

01

Draw the Inequality Constraints on a Graph

First, we will plot each inequality on a graph. To do this, treat each inequality like an equation, and then shade the region satisfying the inequality: 1. \(\frac{x}{4}\leq y\) becomes y = \(\frac{x}{4}\), shade above the line. 2. \(y \leq \frac{2x}{3}\) becomes y = \(\frac{2x}{3}\); shade below the line. 3. \(x+y \geq 5\) becomes y = -x + 5; shade above the line. 4. \(x+2y \leq 10\) becomes y = \(\frac{1}{2}\)(10 - x); shade below the line. 5. \(x \geq 0\), shade right of the y-axis. 6. \(y \geq 0\), shade above the x-axis. Please note that we're working with x and y greater than or equal to 0, so our graph will be in the first quadrant.
02

Identify Feasible Region

The feasible region is the area in which all shaded regions overlap. In this case, it is a quadrilateral formed by the intersection of the lines \(y = \frac{x}{4}\), \(y=\frac{2x}{3}\), \(y = -x + 5\), and \(y = \frac{1}{2}(10-x)\).
03

Identify the Corner Points

The corner points are the vertices of the quadrilateral we identified as the feasible region. These are the points where two constraints intersect. Find these points by solving the corresponding equation pairs: 1. Intersection of \(y = \frac{x}{4}\) and \(y = \frac{1}{2}(10 - x)\): (2, 0.5) 2. Intersection of \(y = -x + 5\) and \(y = \frac{1}{2}(10 - x)\): (2, 3) 3. Intersection of \(y = -x + 5\) and \(y = \frac{2x}{3}\): (3, 2) 4. Intersection of \(y = \frac{x}{4}\) and \(y = \frac{2x}{3}\): (0, 0)
04

Evaluate Objective Function at Corner Points

Plug in the x and y coordinates of each corner point in the objective function c = 3x - 3y, and see which coordinates minimize the objective function: 1. c(2, 0.5) = 3(2) - 3(0.5) = 6 - 1.5 = 4.5 2. c(2, 3) = 3(2) - 3(3) = 6 - 9 = -3 3. c(3, 2) = 3(3) - 3(2) = 9 - 6 = 3 4. c(0, 0) = 3(0) - 3(0) = 0
05

Determine the Minimum Value

The smallest value among the evaluated corner points is when c = -3 at the point (2, 3). So, the optimal solution exists, and the minimum value of the objective function is -3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Objective Function Optimization
Objective Function Optimization in linear programming deals with finding the best possible outcome for a linear equation, usually in terms of minimizing or maximizing a particular value. In this context, the objective function is given as \( c = 3x - 3y \). When we say that we want to "minimize" this function, we are looking for the values of \( x \) and \( y \) that make \( c \) as small as possible.

To solve this, we plug in the values of \( x \) and \( y \) from specific points, known as corner points, into the function \( c = 3x - 3y \). These corner points are derived from the feasible region, where all constraints are satisfied. By evaluating each point, we determine which coordinates minimize the function. For instance, by calculating \( c \) at each corner, we can see how the output changes and identify the smallest \( c \).

This procedure helps in determining not just any solution, but the optimal solution that leads to the minimal value of \( c \), which in this case is -3, occurring at the point (2, 3).
Inequality Constraints
Inequality Constraints are a fundamental part of linear programming and serve as the boundaries within which the solution must lie. For this problem, the constraints are a set of inequalities involving \( x \) and \( y \):

  • \( \frac{x}{4} \leq y \)
  • \( y \leq \frac{2x}{3} \)
  • \( x + y \geq 5 \)
  • \( x + 2y \leq 10 \)
  • \( x \geq 0 \)
  • \( y \geq 0 \)

These inequalities describe relationships between \( x \) and \( y \), establishing limits within which solutions can exist. Graphically, each inequality represents a region on the Cartesian plane, and only the area where all these conditions meet is considered feasible.

To interpret these constraints, you graph each inequality and carefully shade the region that satisfies each one. The challenges often involve ensuring this overlap is properly identified, as the final solution is contingent on adhering strictly to these constraints.

Accurate representation of these boundaries ensures that all calculations and outcomes will respect the given limits, with viable solutions situated only within this prescribed space.
Feasible Region Identification
Feasible Region Identification is a key step in solving linear programming problems and refers to the area where all constraint inequalities intersect, the region where a viable solution might exist.

To determine this region, you first draw the lines representing each inequality on a coordinate plane. Then, you identify the intersection or overlap of all these different shaded regions. In our example, the feasible region is a quadrilateral, formed by the lines \( y = \frac{x}{4} \), \( y = \frac{2x}{3} \), \( y = -x + 5 \), and \( y = \frac{1}{2}(10-x) \).

This region must also satisfy all non-negativity constraints (\( x \geq 0 \), \( y \geq 0 \)), meaning it lies entirely within the first quadrant of the graph. By finding where these lines cross, we define the vertices or corner points of the feasible region, which is crucial in determining the optimal solution. Each vertex is a potential candidate for optimizing the objective function.

The feasible region essentially sets the stage for where solutions can exist, guiding the process of finding maximum or minimum values for the objective function based on the given constraints.

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Most popular questions from this chapter

The Scottsville Textile Mill produces several different fabrics on eight dobby looms which operate 24 hours per day and are scheduled for 30 days in the coming month. The Scottsville Textile Mill will produce only Fabric 1 and Fabric 2 during the coming month. Each dobby loom can turn out \(4.63\) yards of either fabric per hour. Assume that there is a monthly demand of 16,000 yards of Fabric 1 and 12,000 yards of Fabric 2. Profits are calculated as 33 d per yard for each fabric produced on the dobby looms. a. Will it be possible to satisfy total demand? b. In the event that total demand is not satisfied, the Scottsville Textile Mill will need to purchase the fabrics from another mill to make up the shortfall. Its profits on resold fabrics ordered from another mill amount to \(20 \mathrm{~d}\) per yard for Fabric 1 and \(16 \mathrm{e}\) per yard for Fabric \(2 .\) How many yards of each fabric should it produce to maximize profits?

$$ \begin{aligned} \text { Minimize } & c=s+t+u \\ \text { subject to } & 3 s+2 t+u \geq 60 \\ & 2 s+t+3 u \geq 60 \\ & s+3 t+2 u \geq 60 \\ & s \geq 0, t \geq 0, u \geq 0 . \end{aligned} $$

$$ \begin{array}{ll} \text { Minimize } & c=s+t+2 u \\ \text { subject to } & s+2 t+2 u \geq 60 \\ & 2 s+t+3 u \geq 60 \\ & s+3 t+6 u \geq 60 \\ & s \geq 0, t \geq 0, u \geq 0 . \end{array} $$

\(\nabla\) \mathrm{\\{} T r a n s p o r t a t i o n ~ S c h e d u l i n g ~ W e ~ r e t u r n ~ t o ~ y o u r ~ e x p l o i t s ~ c o - ~ ordinating distribution for the Tubular Ride Boogie Board Company. \({ }^{36}\) You will recall that the company has manufacturing plants in Tucson, Arizona and Toronto, Ontario, and you have been given the job of coordinating distribution of their latest model, the Gladiator, to their outlets in Honolulu and Venice Beach. The Tucson plant can manufacture up to 620 boards per week, while the Toronto plant, beset by labor disputes, can produce no more than 410 Gladiator boards per week. The outlet in Honolulu orders 500 Gladiator boards per week, while Venice Beach orders 530 boards per week. Transportation costs are as follows: Tucson to Honolulu: \(\$ 10 /\) board; Tucson to Venice Beach: \(\$ 5 /\) board; Toronto to Honolulu: \(\$ 20 /\) board; Toronto to Venice Beach: \(\$ 10 /\) board. Your manager has said that you are to be sure to fill all orders and ship the boogie boards at a minimum total transportation cost. How will you do it?

Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. \(\begin{aligned} \text { Maximize } & p=2 x+3 y \\ \text { subject to } & 0.1 x+0.2 y \geq 1 \\ & 2 x+\quad y \geq 10 \\ & x \geq 0, y \geq 0 . \end{aligned}\)

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