Math 3A Meadows or Malls? Review

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Math 3A Meadows or Malls? Review Name Linear Programming w/o Graphing (2 variables) 1. A manufacturer makes digital watches and analogue (non-digital) watches. It cost $15 to make digital watch and $20 to make an analogue watch. The manufacturer can spend no more than $3000 making both types of watches each day. The company needs to make no fewer than 50 digital watches each day. Since they market to many young kids, the company wants the number of digital watches produced to be at least twice the number analogue watches produced. The company stands to earn $6 for each digital watch and $9 for each analogue watch. How many of each type of watch should be produced each day in order to maximize its profits? This should be solved algebraically. Show all work and fill out chart below. Variables: d: # of digital watches made, a: # of analogue watches made. Constraints: A. (money) B. (no fewer than 50 digital) C. (digital at least 2x analogue.) D. (don t make negative watches. You don t need a similar constraint for d (why?).) If you have trouble deciding which number to put the 2 on, test some values. If the manufacturer makes 100 d and 100 a (not allowed as per the wording in this constraint), then this will state, which is false. If they make 200 d and 100 a, this will state which is true in this constraint. Your initial instinct might be to put the 2 on the d, because you are making more digital watches, but you actually need to multiply the one you are making less of to balance that out) Since there are only two variables, make PAIRS of constraints to test. Combination Solution ( d, a ) In feasible region? (test against all other constraints) AB No (breaks C) AC (120, 60) Yes 1260 AD (200,0) Yes 1200 BC (50,25) Yes 525 BD (50,0) Yes 300 CD (0,0) No (breaks B) Profit 6d+9a WORK ON NEXT PAGE FOR TABLE. When you are listing the combinations, you can be sure you have them all if you think about alphabetizing the combinations. List in order, and don t list non-alphabetical combinations (i.e. don t list AB because it s not in alpha order.) This avoids duplicates and makes sure you get them all.

When finding potential corner points, convert the constraints from inequalities to equations if needed. A. B. C. D. AB: Solution: AC: Solution: (120, 60) AD: Solution: (200,0) BC: Solution: (50,25) BD: Solution: (50,0) CD: Solution (0,0)

Be able to solve systems using graphing, substitution, and elimination. 2. 2 f 4 f 3g 5g 40 30 Coefficients with common factors might lead you to choose to eliminate f. Multiply the top equation by -2 and rewrite: Back substitute Solution: 3. 5w 4z 12 7.5w 6z 20 I notice that 7.5 x 2 is 15, and that 3x5 is 15. I multiple the top equation by -3, and the bottom by 2 and rewrite. Since this is an equation that is FALSE, there are no solutions to this system. This might also be a good system to solve by graphing. If you did, you would notice that the equations have the same slope but different y-intercepts (i.e. are parallel)

4. 2r 3s t 3 r 2s 4t 2 4r s 7t 8 The most useful term here is the t in the first equation. 1 coefficients are always easy to multiply, and the sign is already opposite of the other t s, which means no dealing with multiplying by negatives. I choose to eliminate t first, by combining equations 1 and 2 and then 1 and 3. Equations 1 and 2: Multiply top equation by 4 (save this for later.) Equations 1 and 3: Multiply top equation by 7: Combining the results: Hey, wait! What if we multiply the top equation by 2. We get This cannot be! This system also has no solutions.

5. 4x 2y 2z 10 2x 8y 4z 32 30x 12y 4z 24 OK. The first thing to notice here is that every equation has all even coefficients. It will help us see the path towards the solution better if we take out all common factors from each row. OK, that s better. Now I can look at this and see that z will be good to eliminate, because of the small coefficients and matching signs. I want to combine equations 2 and 3 for sure, and then I ll also do equations 1 and 2 because the signs are alternating. equations 2 and 3: Or equivalently: Equations 1 and 2: Multiply top by 2 (yes, this brings us back to the original equation 1. A little bit of planning could have avoided this step.) Combine the results: No common factors, so let s multiply the top by 6 and the bottom by -5, with the goal of canceling y. Back substitute:

Solution:

Solve the following problems using systems of equations. 5. A computer store sells two different types of computers, A and B. Computer A costs $2000 and Computer B costs $2200. If a total of 27 computers are sold and $56,800 is collected from these sales, how many Type A computers are sold? Solve by elimination: multiply top by -2000. (# sold) (Sales) Then, since: The store sold 13 of computer A and 14 of computer B 6. Mike and Melissa went to the grocery store. Mike bought 5 apples and 6 bags of chips. Melissa bought 3 apples and 3 bags of chips. Mike spent $12.30 and Melissa spent $6.30. If Layna goes to the same store and buys 2 apples and 1 bag of chips, how much will her purchase cost? Show all work and explain how you know your answer is correct. If Melissa doubled her order, she would have 6 apples and 6 chips for $12.60. She would also have exactly one more apple than Mike, and thus conclude that the apple costs $0.30. Her three bags of chips cost $.90, out of her total order of $6.30. That means she spent $5. 40 on the three bags of chips. That means each bag of chips cost $1.80. Layna buys 2 30 cent apples and 1 $1.80 bag of chips. Her price will be $2.40. 7. The senior classes at High School A and High School B planned separate trips to New York City. The senior class at High School A rented and filled 1 van and 6 buses with 372 students. High School B rented and filled 4 vans and 12 buses with 780 students. Each van and each bus carried the same number of students. How many students can a van carry? How many students can a bus carry? Let v be the number of students a van can carry and b be the number of students a bus can carry. Multiply top by 2: Substitute:

A van can carry 18 students, and a bus can carry 59 students. (don t make the mistake of saying 18 vans and 59 buses go back to the place where you defined your variables.) 8. Enrique and Ryan are selling fruit for a school fundraiser. Customers can buy small boxes of oranges and large boxes of oranges. Enrique sold 3 small boxes of oranges and 14 large boxes of oranges for a total of $203. Ryan sold 11 small boxes of oranges and 11 large boxes of oranges for a total of $220. Find the cost of one small box of oranges and one large box of oranges. S: cost of a small box L: cost of a large box What gross numbers. Wait! Try dividing the bottom equation by 11. Then Multiply bottom by -3 Since, and then A small box of fruit costs $7, and a large box costs $13.

Write the constraints and solve the following problem. 9. Joe is having a fundraiser for a trip to go to Willy Wonka s Chocolate Factory. He plans to sell cans of soda and chocolate bars at school to raise the money. He is selling the soda for $1 and the chocolate for $1. His uncle gave him $100 to spend to buy his supplies to sell. Cans of soda cost $0.50 and chocolate bars cost $0.75. He can carry at most 100 sodas and 100 chocolate bars in his bookbag. He wants to sell at least twice as many chocolate bars as cans of soda because he wants to share the joy of chocolate with the world. How many sodas and chocolate bars should he buy to sell to maximize his profit? A. (purchasing supplies) B. (bookbag size) C. (bookbag size) D. (he loves chocolate!) (remember the coefficient goes on the number he wants to sell less of to keep everything balanced) Profit: (have to subtract the purchase price from the selling price) Combination testing: Let s do a few things here. Rewrite constraints with equals signs, AND multiply constraint A by 100 to get rid of decimals. A. B. C. D. AB: AC AD

BC BD CD OK, make a table! Combination Solution (s,c) In feasible region? (test against all other constraints) AB No (breaks D) Profit.5s+.25c AC (66.6, 100) No (breaks D) AD (57.14, 114.29) No (breaks C) BC (100,100) No (breaks D) BD (100,200) No (breaks C) CD (50,100) Yes! (checked against A, just to be sure.) A. B. C. D. (recopy list of constraints for check step.) Joe should purchase 50 cans of soda and 100 chocolate bars to sell in order to raise money for his trip.

10. Given the constraints for a situation, find all combinations, solve each combination, identify the corner points and find the solution that will maximize the profit. l c t 20 l 3 c 2t 24 c 0 t 0 Pr ofit 0.5l 0.75 c.6t These constraints should be familiar to you: they are the same as in Ming s New Maneuver. Here is the book s list of constraints, and the book s table of results. A few things to ponder: Why didn t we list constraint IV like the book does? Why does the book ONLY list combinations that include equation I? Test the four remaining values in the profit equation to find the best result in this problem.

11. Name a point that is behind the xy plane and to the left of the yz plane. (-2, 5, 12) ( -, any, +) 12. Name a point that is beneath the xz plane and to the right of the yz plane. (2, -5, 10) (+, -, any) 13. Name a point that is above the xz plane and behind the xy plane. (10, 10, -5) (any, +, -) 14. What is the difference between the graph of an equation with 2 variables and the graph of an equation with 3 variables? A three variable equation s graph is a plane in 3d space Wrinkle: even if one or more variable is zero, in 3d space read it like is still a plane, since you should A two variable equation s graph is a line in 2d space. (even the graph like is a vertical line, since it s