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\(\begin{array}{lc}\text { Maximize } & p=12 x+10 y \\ \text { subject to } & x+y \leq 25 \\ & x \quad \geq 10 \\ & -x+2 y \geq 0 \\ x & \geq 0, y \geq 0 .\end{array}\)

Short Answer

Expert verified
The optimal solution for the given linear programming problem is at the vertex \((20/3, 35/3)\) with a maximum profit of p = 290.

Step by step solution

01

Graph the Constraints

First, we will plot the given constraint inequalities on a Cartesian plane: 1. \(x + y \leq 25\) 2. \(x \geq 10\) 3. \(-x + 2y \geq 0\) 4. \(x \geq 0\) 5. \(y \geq 0\) To make it easier to graph, we can rewrite the inequalities as equations: 1. \(x + y = 25\) 2. \(x = 10\) 3. \(y = (1/2)x\) Graph these equations and shade the appropriate regions of each inequality. The overlapping, shaded region will be the feasible region.
02

Identify the Vertices

Next, we need to find the vertices of the feasible region by solving the systems of two equalities corresponding to the constraint boundaries. 1. Intersection of \(x + y = 25\) and \(x = 10\): Solve for x and y to get the point (10,15). 2. Intersection of \(x + y = 25\) and \(y = (1/2)x\): Solve for x and y to get the point (20/3, 35/3). 3. Intersection of \(x = 10\) and \(y = (1/2)x\): Solve for x and y to get the point (10, 5).
03

Compute the Profit at Each Vertex

Now we will compute the profit function at each vertex of the feasible region: 1. Profit at (10, 15): \(p = 12(10) + 10(15) = 270\) 2. Profit at (20/3, 35/3): \(p = 12(20/3) + 10(35/3) = 290\) 3. Profit at (10, 5): \(p = 12(10) + 10(5) = 170\)
04

Identify the Optimal Solution

The vertex with the maximum profit is (20/3, 35/3) with a profit of 290. Therefore, the optimal solution is \(x = 20/3\) and \(y = 35/3\) with a maximum profit of p = 290.

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Key Concepts

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

Optimization Problems
Optimization problems are mathematical questions aimed at finding the best solution from all feasible solutions. In the context of linear programming, they involve maximizing or minimizing a linear objective function, which represents a certain quantity that needs to be optimized – such as profit, cost, or distance. The optimization is subject to a set of linear inequalities known as constraints that define the feasible region in which the optimum solution lies.

For example, consider the problem of maximizing the profit function, given by the equation \(p = 12x + 10y\), subject to certain constraints like \(x + y \textless{}= 25\), \(x \textgreater{}= 10\), and \(y \textgreater{}= 0\). The aim is to determine the quantity of two products, \(x\) and \(y\), that a company should produce to achieve the maximum profit without violating any constraints imposed by resources or market demands.
Graphical Method
The graphical method is a technique to solve linear programming problems involving two variables. It involves graphing the constraints on a two-dimensional plane and identifying the region that satisfies all constraints, known as the feasible region. Once the feasible region is established, the objective function is evaluated at the corners or vertices of this region to find the optimum value.

Using the graphical method in our example, you begin by plotting the constraints \(x + y = 25\), \(x = 10\), and \(y = 0.5x\). Then, you determine the overlapping area that satisfies all inequalities. This visual approach is particularly helpful for understanding how different constraints interact and where the potential solution may lie. By examining the vertices of the feasible region, you can quickly compute the value of the objective function at those points and select the optimal solution.
Constraint Inequalities
Constraint inequalities are equations or inequalities that define the conditions or limitations within which an optimization problem must be solved. They form the boundary of the feasible region and typically represent limitations such as resource availability, production capacities, or market requirements. The solutions to an optimization problem must not only optimize the objective function but also fit within these constraints.

In our exercise example, the constraints are given by the inequalities \(x+y \textless{}= 25\), \(x \textgreater{}= 10\), and \(-x + 2y \textgreater{}= 0\), along with \(x \textgreater{}= 0\) and \(y \textgreater{}= 0\) to ensure non-negative production quantities. These inequalities shape the feasible region and play a crucial role in determining where the maximum or minimum value of the objective function can be found.
Feasible Region
The feasible region is the set of all possible points that satisfy the constraint inequalities of a linear programming problem. It represents all the potential solutions that adhere to the problem's limitations. The feasible region is key to finding the optimal solution because this solution must be among the points within this region.

In our example involving profit maximization, the feasible region is graphically depicted as the area on the Cartesian plane that is bounded by the lines corresponding to the constraints and lies above the x and y axes. To find the optimal point, you look at the vertices of this region, as the maximum or minimum values for a linear objective function in a bounded region will always be found at one of these vertices. This is because the linear function will increase or decrease uniformly within this region, therefore the extremities will provide the optimal values.

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Most popular questions from this chapter

The Megabuck Hospital Corp. is to build a state-subsidized nursing home catering to homeless patients as well as high-income patients. State regulations require that every subsidized nursing home must house a minimum of 1,000 homeless patients and no more than 750 high-income patients in order to qualify for state subsidies. The overall capacity of the hospital is to be 2,100 patients. The board of directors, under pressure from a neighborhood group, insists that the number of homeless patients should not exceed twice the number of high-income patients. Due to the state subsidy, the hospital will make an average profit of \(\$ 10,000\) per month for every homeless patient it houses, whereas the profit per high-income patient is estimated at \(\$ 8,000\) per month. How many of each type of patient should it house in order to maximize profit? HINT [See Example 3.]

Resource Allocation Meow makes cat food out of fish and cornmeal. Fish has 8 grams of protein and 4 grams of fat per ounce, and cornmeal has 4 grams of protein and 8 grams of fat. A jumbo can of cat food must contain at least 48 grams of protein and 48 grams of fat. If fish and cornmeal both cost 5elounce, how many ounces of each should Meow use in each can of cat food to minimize costs? What are the shadow costs of protein and of fat? HINT [See Example 2.]

f Purchasing Federal Rent-a-Car is putting together a new fleet. It is considering package offers from three car manufacturers. Fred Motors is offering 5 small cars, 5 medium cars, and 10 large cars for \(\$ 500,000\). Admiral Motors is offering 5 small, 10 medium, and 5 large cars for \(\$ 400,000\). Chrysalis is offering 10 small, 5 medium, and 5 large cars for \(\$ 300,000\). Federal would like to buy at least 550 small cars, at least 500 medium cars, and at least 550 large cars. How many packages should it buy from each car maker to keep the total cost as small as possible? What will be the total cost?

Describe at least one drawback to using the graphical method to solve a linear programming problem arising from a real-life situation.

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

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