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Chapter 8: Problem 26

Find the maximum or minimum value of each objective function subject to the given constraints. Maximize \(R(x, y)=50 x+20 y\) subject to \(x \geq 0, y \geq 0,3 x+y \leq 18,\) and \(2 x+y \leq 14\)

Short Answer

Expert verified
The maximum value is 320 at the point (4, 6).

Step by step solution

01

- Identify the Objective Function

The objective function to maximize is given by:\[R(x, y) = 50x + 20y\]
02

- Write the Constraints

The given constraints for the variables are:\(x \geq 0\)\(y \geq 0\)\[3x + y \leq 18\]\[2x + y \leq 14\]
03

- Graph the Constraints

Plot the lines of the inequalities on a graph. First, rearrange the inequalities into slope-intercept form:\[3x + y = 18 \rightarrow y = 18 - 3x\]\[2x + y = 14 \rightarrow y = 14 - 2x\]Then, identify the feasible region which is the intersection of all the inequalities, including the non-negativity constraints, in the first quadrant.
04

- Find the Corner Points

Determine the points at the vertices (corner points) of the feasible region. These can be found by solving the system of equations at the intersections of the constraint lines:Intersection of \(y = 18 - 3x\) and \(y = 14 - 2x\).Set \(18 - 3x = 14 - 2x\), solving gives \(x = 4\), \(y = 6\).Other corner points are where the lines intersect the x and y axes: \((0, 14)\) and \((6, 0)\) respectively.
05

- Evaluate the Objective Function

Substitute these corner points into the objective function to evaluate:1. \(R(0, 14) = 50(0) + 20(14) = 280\)2. \(R(4, 6) = 50(4) + 20(6) = 320\)3. \(R(6, 0) = 50(6) + 20(0) = 300\)Compare these values to find the maximum.
06

- State the Maximum Value

The maximum value of the objective function \(R(x, y)\) is 320, occurring at the point \((4, 6)\).

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Key Concepts

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

Objective Function
In linear programming, the objective function is the formula you want to maximize or minimize. For our exercise, the objective function is defined as:\[R(x, y) = 50x + 20y\]This function represents what we are trying to achieve, which in this case, is to find the maximum value of \(R(x, y)\). The coefficients (50 and 20) show the weight or contribution of each variable (\(x\) and \(y\)) towards achieving that goal.
Constraints
Constraints are the conditions you need to satisfy for your solution to be valid. In our problem, the constraints are:
  • \(x \, \geq \, 0\)
  • \(y \, \geq \, 0\)
  • \[3x + y \, \leq \, 18\]
  • \[2x + y \, \leq \, 14\]
These inequalities limit the values \(x\) and \(y\) can take. They define the boundaries within which we need to find our objective function's maximum or minimum.
Feasible Region
The feasible region is the shaded area on a graph where all the constraints overlap. To identify it, we first convert the constraints equations into their line equations:\[3x + y = 18 \, \text{(rearranged to)} \, y = 18 - 3x\]\[2x + y = 14 \, \text{(rearranged to)} \, y = 14 - 2x\]Next, graph these lines along with the non-negativity constraints \(x \, \geq\, 0\) and \(y \, \geq\, 0\). The feasible region will be the intersection area of all these lines in the first quadrant of the graph.
Corner Points
Corner points are the vertices of the feasible region. To find these, we locate where the constraint lines intersect.For this exercise:
  • Intersection of \(y = 18 - 3x\) and \(y = 14 - 2x\): Set \(18 - 3x = 14 - 2x\), solving gives \(x = 4\), \(y = 6\).
  • Intersection with the axes: The other points are \((0, 14)\) and \((6, 0)\).
These points are critical because the maximum or minimum value of the objective function always lies at one of these points.
Maximum Value
To determine the maximum value of the objective function, substitute each corner point back into \(R(x, y)\). For our example:
  • At \((0, 14)\):\[R(0, 14) = 50(0) + 20(14) = 280\]
  • At \((4, 6)\):\[R(4, 6) = 50(4) + 20(6) = 320\]
  • At \((6, 0)\):\[R(6, 0) = 50(6) + 20(0) = 300\]
Compare these values to find that the maximum value of the objective function is 320, occurring at the point \((4, 6)\).

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Most popular questions from this chapter

Solve each problem by using a system of three linear equations in three variables. Milk, Coffee, and Doughmuts The employees from maintenance go for coffee together every day at 9 A.M. On Monday, Hector paid \(\$ 5.45\) for three cartons of milk, four cups of coffee, and seven doughnuts. On Tuesday, Guillermo paid \(\$ 5.30\) for four milks, two coffees, and eight doughnuts, On Wednesday, Anna paid \(\$ 5.15\) for two milks, five coffees, and six doughnuts. On Thursday, Alphonse had to pay for five milks, two coffees, and nine doughnuts. How much change did he get back from his \(\$ 10\) bill?

Solve each problem using two variables and a system of two equations. Solve the system by the method of your choice. Note that some of these problems lead to dependent or inconsistent systems. One plan for federal income tax reform is to tax =an individual's income in excess of \(\$ 15,000\) at a \(17 \%\) rate. Another plan is to institute a national retail sales tax of \(15 \% .\) If =an individual spends \(75 \%\) of his or her income in retail stores where it is taxed at \(15 \%\), then for what income would the =amount of tax be the same under either plan?

Solve the system \(5 x-9 y=12\) and \(18 y-10 x=20\).

Solve each problem using two variables and a system of two equations. Solve the system by the method of your choice. Note that some of these problems lead to dependent or inconsistent systems. Doctors often prescribe the same drugs for children as they do for adults. If \(a\) is the age of a child and \(D\) is the adult dosage, then to find the child's dosage \(d\) doctors can use the formula \(d=0.08 a D\) (Fried's rule) or \(d=D(a+1) / 24\) (Cowling's rule). For what age do the two formulas give the same child's dosage?

Solve each system of inequalities. $$\begin{aligned}&x^{2}+y^{2} \geq 9\\\&x^{2}+y^{2} \leq 25\\\&y \geq|x|\end{aligned}$$
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