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$$ \begin{array}{ll} \text { Minimize } & c=2 s+t \\ \text { subject to } & 3 s+t \geq 30 \\ & s+t \geq 20 \\ & s+3 t \geq 30 \\ & s \geq 0, t \geq 0 \end{array} $$

Short Answer

Expert verified
The minimum value of the objective function c = 2s + t under the given constraints is \(c = 25\), and it occurs at the point \((s,t) = (9,7)\).

Step by step solution

01

Plot the feasible region

First, we need to plot the constraints on the coordinate plane: 1. \(3s + t \geq 30\) 2. \(s + t \geq 20\) 3. \(s + 3t \geq 30\) 4. \(s \geq 0, t \geq 0\) Take each inequality and solve it for t to help with graphing them: 1. \(t \geq 30 - 3s\) 2. \(t \geq 20 - s\) 3. \(t \geq \frac{30 - s}{3}\) Plot these inequalities and shade the feasible region - the area that satisfies all the inequalities.
02

Identify the vertices of the feasible region

Once the feasible region is plotted, we need to find the vertices by finding the intersection points of the constraint lines. The vertices are the points where any combination of two inequality lines intersects. To find these intersection points, we'll formulate a system of two linear equations for each possible combination of two constraints: 1. Intersection of \(3s + t = 30\) and \(s + t = 20\): \(\begin{cases} 3s + t = 30\\ s + t = 20 \end{cases} \) 2. Intersection of \(3s + t = 30\) and \(s + 3t = 30\): \(\begin{cases} 3s + t = 30\\ s + 3t = 30 \end{cases} \) 3. Intersection of \(s + t = 20\) and \(s + 3t = 30\): \(\begin{cases} s + t = 20\\ s + 3t = 30 \end{cases} \) Solve these systems of equations: 1. Intersection point: \((10, 10)\) 2. Intersection point: \((9, 7)\) 3. Intersection point: \((6, 14)\) These are the vertices of the feasible region.
03

Evaluate the objective function at each vertex

Now, we need to evaluate the objective function (c = 2s + t) at each vertex to determine the minimum value of c. 1. For the vertex \((10,10)\): \(c = 2(10) + 10 = 30\) 2. For the vertex \((9,7)\): \(c = 2(9) + 7 = 25\) 3. For the vertex \((6,14)\): \(c = 2(6) + 14 = 26\)
04

Determine the minimum value and the corresponding vertex

Comparing the values of c at each vertex, we see that the minimum value of c is 25, and it occurs at the vertex (9,7). Thus, the minimum value of the objective function is c = 25, and it happens at the point (s,t) = (9,7).

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Key Concepts

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

Feasible Region
In the world of linear programming, the feasible region is like the playground where all the solutions live. It represents all the possible solutions that satisfy the given inequality constraints. To visualize this, we usually plot the constraints on a coordinate plane. Here, the inequalities are turned into boundaries that form a region which we call the feasible region.

When we plot inequalities, we often shade the area where all these conditions are true. The feasible region lies where all shaded areas overlap. This is a crucial step, as this region essentially contains all the points that could potentially solve our optimization problem.

For our given exercise, the boundary is formed by the constraints \(3s + t \geq 30\), \(s + t \geq 20\), and \(s + 3t \geq 30\), alongside the non-negativity constraints \(s \geq 0\) and \(t \geq 0\). After plotting these, you will notice a certain area where all conditions meet. This is the feasible region. Every point inside this region is a valid solution to the linear programming problem.
Objective Function
The objective function in a linear programming problem is what you are trying to optimize. This could either be maximizing or minimizing a particular value. In this exercise, we are trying to minimize our objective function, given by \(c = 2s + t\).

The essence of solving such a problem lies in evaluating this function within the feasible region. To find the best possible value, we calculate the function's value at each vertex of the feasible region. Why the vertices? Because of the linear nature of our problem, the minimum or maximum value will always lie at the edges of our feasible region, typically at a corner (vertex).
  • Evaluate the function at each vertex.
  • Compare to find the minimum or maximum value.
For our problem, the objective function is evaluated at the points (10,10), (9,7), and (6,14). By comparing these values, we identify the vertex that gives us the minimum value. Thus, knowing and understanding the objective function is key to solving linear programming problems.
Inequality Constraints
Inequality constraints define the limits or the boundaries of your feasible region. They are the conditions that your solutions must satisfy. In linear programming, these constraints are usually linear inequalities. Each of these inequalities represents a half-space on the coordinate plane.

By plotting each inequality as an equation, you get a line that serves as the boundary. For example, in our exercise, the constraint \(3s + t \geq 30\) becomes the boundary line \(3s + t = 30\). Similarly, the other constraints do the same.
  • Each inequality splits the plane into two halves.
  • The feasible region is where all these halves intersect.
This ensures that all the conditions are met simultaneously. Without these constraints, the problem would not have a limited number of solutions. In fact, it might not even have any solutions! That's why inequality constraints play a fundamental role in defining the problem's scope and ensuring feasible solutions can be found within the specified conditions.

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Most popular questions from this chapter

Management \(^{20}\) You are the service manager for a supplier of closed- circuit television systems. Your company can provide up to 160 hours per week of technical service for your customers, although the demand for technical service far exceeds this amount. As a result, you have been asked to develop a model to allocate service technicians' time between new customers (those still covered by service contracts) and old customers (whose service contracts have expired). To ensure that new customers are satisfied with your company's service, the sales department has instituted a policy that at least 100 hours per week be allocated to servicing new customers. At the same time, your superiors have informed you that the company expects your department to generate at least \(\$ 1,200\) per week in revenues. Technical service time for new customers generates an average of \(\$ 10\) per hour (because much of the service is still under warranty) and for old customers generates \(\$ 30\) per hour. How many hours per week should you allocate to each type of customer to generate the most revenue?

Finance Senator Porkbarrel habitually overdraws his three bank accounts, at the Congressional Integrity Bank, Citizens' Trust, and Checks R Us. There are no penalties because the overdrafts are subsidized by the taxpayer. The Senate Ethics Committee tends to let slide irregular banking activities as long as they are not flagrant. At the moment (due to Congress" preoccupation with a Supreme Court nominee), a total overdraft of up to \(\$ 10,000\) will be overlooked. Porkbarrel's conscience makes him hesitate to overdraw accounts at banks whose names include expressions like "integrity" and "citizens' trust." The effect is that his overdrafts at the first two banks combined amount to no more than one-quarter of the total. On the other hand, the financial officers at Integrity Bank, aware that Senator Porkbarrel is a member of the Senate Banking Committee, "suggest" that he overdraw at least \(\$ 2,500\) from their bank. Find the amount he should overdraw from each bank in order to avoid investigation by the Ethics Committee and overdraw his account at Integrity by as much as his sense of guilt will allow.

I Purchasing Cheapskate Electronics Store needs to update its inventory of stereos, TVs, and DVD players. There are three suppliers it can buy from: Nadir offers a bundle consisting of 5 stereos, 10 TVs, and 15 DVD players for \(\$ 3,000\). Blunt offers a bundle consisting of 10 stereos, \(10 \mathrm{TVs}\), and 10 DVD players for \(\$ 4,000\). Sonny offers a bundle consisting of 15 stereos, \(10 \mathrm{TVs}\), and 10 DVD players for \(\$ 5,000\). Cheapskate Electronics needs at least 150 stereos, \(200 \mathrm{TVs}\), and 150 DVD players. How can it update its inventory at the least possible cost? What is the least possible cost?

$$ \begin{array}{ll} \text { Maximize } & p=x+y+z+w \\ \text { subject to } & x+y+z \leq 3 \\ & y+z+w \leq 3 \\ & x+z+w \leq 4 \\ & x+y+w \leq 4 \\ & x \geq 0, y \geq 0, z \geq 0, w \geq 0 . \end{array} $$

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