Math 119/211 Test 2 – Surgent – Spring 2005 – Ch 9-11



Math 211 Test 2 Review problems

Chapter 4

1. Determine the feasible region and all corner points of the system of inequalities:

[pic]

Feasible Region

Corner points are [pic]

2. Minimize [pic] subject to [pic]. Use linear programming. Clearly mark the feasible region and the corner points.

Feasible region

Or solution set

Corner points and z value:

(0, 4), z = 108 (5, 2), z = 104 min, (11, 0), z = 110

3. You are thinking of making your mansion more energy efficient by replacing some of the light bulbs with compact fluorescent bulbs, and installing part or all of your exterior walls. Each compact fluorescent light bulb costs $4 and saves you an average of $2 per year in energy costs, and each square foot of wall installation costs $1 and saves you an average of $0.20 per year in energy costs. Your mansion has 200 light fittings and 3000 square feet of uninstalled exterior wall. To impress your friends, you would like to spend as much as possible, but save no more than $800 per year in energy costs (you are proud of your large utility bills). Find how many compact fluorescent light bulbs and how many square feet of installation should you purchase and how much will you save in energy costs per year. Clearly write all the variables, constraint inequalities and objective function showing feasibility region and all corner points.

Solution: Variables: x is the number of bulbs and y is sq. ft of installation

Cost C = 4x + y

Subject to C(0, 3000) A

[pic] Feasible region B

D(200, 0)

The other points: A(100, 3000), and B(200, 2000)

Check that at the point A, cost is $3400, savings $800 for 100 bulbs and 3000 sq. ft of installation, which is the optimum.

4. Find the value of K so that the given system has no solution:

[pic]

Solution: Find the determinant value of the coefficient matrix and set equal to zero.

[pic]

Using Sarrus’s rule see that [pic]

5. Voting for an award went like this: Ken had 7 first-place votes, 2 second-place votes and 1 third-place vote, for a total of 43 points. Jane had 5 first-place votes, 2 second-place votes and 3 third-place votes for 37 points. Lou had 1 first-place vote, 4 second-place votes and 5 third-place votes for 27 points.

If you had 3 first-place votes, 4 second-place votes and 3 third-place votes, what would have been your point total? You may apply any method you know for the solution.

Solution:

[pic]

The total point value is [pic]

6. From the given matrices find X when[pic].

[pic]

Solution:

[pic]

7. Let [pic]. Find and simplify 3A + 2C.

Solution:

[pic]

8. Using Augmented Matrices and Row Operations solve for x, y and z

[pic]

Solution:

[pic]

9. Given that [pic], then find m, a and b if [pic]

Solution: [pic]

Write answer here: m = 4 a = 6 b = 2

Evaluate the determinant of the 3 by 3 matrix

[pic] you may use Sarrus rule also

10. Use Shortcut for 2 by 2 matrix

Let [pic]then [pic]

Example:

[pic]then [pic]

11. Solve (using Cramer’s Rule)

[pic]

[pic] [pic] [pic]

Now, [pic] and [pic]

Practice problems:

1. Solve the system algebraically by elimination: [pic]

2. Solve the system [pic]. If the system of equations is inconsistent or dependent , so state.

3. Use the augmented matrix form to find solution if any: [pic]

4. Use rref to find solution if any:

[pic]

5. Find the value of [pic]for the following matrices:

[pic] and [pic]

6. Find the inverse of the matrix [pic]

7. Solve the system [pic] by matrix inverse.

8. Minimize [pic] subject to [pic]. Use linear programming. Clearly mark the feasible region and the corner points.

9. Identify the feasible region and the corner points: [pic]

10. A company manufactures two types of fancy wooden display cabinets. Each cabinet undergoes cutting (of raw materials), sanding and assembly, all of which are handled by machines. Cabinet A requires 2 hours of cutting, 1 hour of sanding and 1 hour of assembly, while cabinet B requires 1 hour of cutting, 1 hour of sanding and 3 hours of assembly. The cutting machine can be used at most 70 hours per week, the sander at most 40 hours and the assembler at most 90 hours. Cabinet A sells at a $40 profit and cabinet B sells at a $60 profit. How many cabinets A and B should be produced to maximize profit?

Solution: Objective function [pic]

With constraints

[pic]

Determine the following corner points (0, 0), (0, 30), (15, 25), (30, 10) and (35, 0) from your feasible region. The maximum is found at (15, 25) which is equal to $2100.

11. Consider this 3 by 3 system:

[pic]

For what value of K this system does not have unique solution? Use Sarrus’s Rule to find K so that the determinant value of the coefficient matrix is zero.

12. Solve for x and y:

13. Each of the matrices given below is in reduced row-echelon form. Determine which matrices have solutions and which have no solution.

• If there is a unique solution, write the solution.

• If there are an infinite number of solutions, write out the solution set.

• If there is no solution, state why.

(a)

(b)

(c)

(d)

14. A factory is producing and selling white and wheat bread. For one loaf of white bread, they need 1 pound of flour and 2 ounces of yeast. For one loaf of wheat bread, they need 1.5 pounds of flour and 1 ounce of yeast. The factory has 1000 pounds of flour and 700 0unces of yeast. White bread sells for $1 per loaf, while wheat bread sells for $1.3 per loaf. Determine the optimal regime of work for the factory: how many loaves of each type of bread should be produced, in order to bring maximal revenue.

Solution: We write F for flour and Y for yeast. x unit of A type white bread and y unit of B type wheat bread. The information is presented below:

| |F |Y |Total |

|White A (x) |1 |2 |1 |

|Wheat B (y) |1.5 |1 |1.3 |

|Availability |1000 |700 | |

To max [pic]

Subject to [pic]

Solving the system we get the intersecting point

(25, 650) and the corner points on the feasible reason

are (0, 666.67), (350, 0), and (0, 0) see the diagram. (0, 0)

The maximum value z = 870 at (25, 650).

|(x, y) |(0, 0) |(350, 0) |(25, 650) |(0, 666.67) |

|z |0 |350 |870 |866.67 |

15. Find the row-reduced echelon form (rref) from the following matrix. You may use calculator. Indicate which function you used with full syntax. Write the solution set.

[pic]

Solution: If we use TI-83 the code is

Entry: 2nd [pic] EDIT

Calculation: 2nd [pic] MATH B (for rref) call your matrix

We have the following result

[pic]

The choice is (D) (1.727, -2.455, -2.818) = (19/11, -27/11, -31/11)

16. Solve the following system using the row-reduced echelon form of the relevant matrices. You may use any features of your calculator. Show your work.

[pic]

Solution: One may use the matrix system in calculator. Edit the matrix and use rref as in example 5 to get the solution (40, 25). The choice is (B) (40, 25)

17. Use inverse matrix method to solve the system

[pic]

Solution: The inverse matrix form is given below as

[pic]

18. Below are row-reduced echelon forms matrices of certain linear systems. For each matrix, tell how many solutions the system has. Explain. If there are any, find the solutions. If there are infinitely many solutions, find the general formula and two particular solutions.

a. [pic] b) [pic] c) [pic]

Solution: a) has unique solution

b. No solution

c. Infinitely many solutions like (t – 1, 2t, 1)

19. Given the matrix [pic]. Find [pic] when [pic]

[pic]

-----------------------

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

(25, 650)

(0, 666.67)

(350, 0)

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