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## Math 1070Q Week 2 Notes

by: Kimberly Clemens

55

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6

# Math 1070Q Week 2 Notes Math 1070Q

Kimberly Clemens
UCONN

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Graphing Linear Inequalities
COURSE
PROF.
Andrew Jaramillo
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Math, Business Math, UCONN, Math for Economics
KARMA
25 ?

## Popular in Applied Mathematics

This 6 page Class Notes was uploaded by Kimberly Clemens on Monday January 11, 2016. The Class Notes belongs to Math 1070Q at University of Connecticut taught by Andrew Jaramillo in Summer 2015. Since its upload, it has received 55 views. For similar materials see Math for Business and Economics in Applied Mathematics at University of Connecticut.

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Date Created: 01/11/16
Math 1070Q Prof. Andrew Jaramillo Week Two Notes Supply, Demand, and Equilibrium Ex 4) The demand equilibrium is p=-2.5x+1692. A firm supplies 50 computers to a market if the price is \$676/computer. The firm will supply 50 more computers if they are priced at \$826/computer. Find the supply equation and the equilibrium point. Step 1) Find the supply equation. X= # of computers, p=price/computer, *p=mx+b *Supply Equation: p=3x+526 Step 2) Find the equilibrium point Set the supply equation equal to the demand equation. 3x+526=-2.5+1692 3x= -2.5x+1166 x=212- This is the amount supplied, or the equilibrium value for demand. To find p, plug x into either supply or demand equation. p=3(212)+526=636+526 p=1162- This is the equilibrium price. *Solution: Eq. Pt. = (212, 1162) Section 3.1- Linear Programming Problems  Understand how linear programming can describe certain scenarios. This section will teach how to optimize (maximize or minimize) a linear function of several variables. An objective function is the function that will be optimized. The quantity of an item will depend on certain factors, like limited time or resources, called constraints. Linear programming is trying to maximize or minimize a function given various constraints. General Strategy for Stating Linear Programming Problems 1. Define the variables. 2. List the objective function and state if it is to be minimized or maximized. This will be an equation. 3. Write “subject to” and list the constraints. This is usually set up in an inequality. Ex 1)*These problems will only be set up and not solved. Manufacturers produce 2 types of boats: dinghies and skiffs. Dinghies make a \$60 profit, requires 2 hours to cut, 4 hours to assemble, and 2 hours to paint. A skiff gives \$80 profit, takes 4 hours to cut, 2 to assemble, and 2 to paint. There is 80 hours of cutting, 84 of assembly, and 50 of painting each week. How many dinghies and skiffs should be produced each week to maximize profit? Write this as a linear programming program. Step 1: define variables- x=# of dinghies, y=#of skiffs Step 2: objective function- the manufacturers want to optimize, in this case specifically maximize, profit. Use price to set up equation. p=60x+80y Step 3: state what factors are subject to constraint. Therefore, state the work of each boat in terms of an inequality and set it to the number allowed for production each week. Cutting: 2x+4y ≤ 80 Assembly: 4x+2y ≤ 84 Painting: 2x+2y ≤ 50 Non-negativity: x ≥ 0, y ≥ 0 because we cannot have “negative” boats. Usually implied in most problems. Ex 2) It costs \$3.75 for a large latte and \$2 for a large hot cocoa. A latte takes 3 minutes to produce, and 2 minutes for hot cocoa. Each day we have 450 large cups and 960 minutes of labor available. How many drinks of each type should be sold to maximize revenue? Write as a linear programming problem. 1: Variablesx= # of lattes, y=# of hot cocoas 2: Objective Function- want the max revenue: R=3.75x+2y 3: Subject to Constraints: time, cups, non negatives. Cups: x+y ≤ 450 Time: 3x+2y ≤ 960 Non negatives: x ≥ 0, y ≥ 0 Ex 3) A farmer has 10 acres to plant purple and orange carrots. Each acre of purple carrots yields 2 tons, each acre of orange yields 4 tons. The farmer wants at least 3 times as many tons of purple carrots as tons of orange carrots. If profits are \$200/ ton of orange carrots, how many acres of each type will maximize profit? 1. Variables: x= # of acres of purple carrots, y=# of acres of orange carrots. tons of purple: 2xtons of orange: 4y 2. We want to maximize profit: p=200(tons of purple)+100(tons of orange) P=200(2x)+100(4y) P=400x+400y 3. Constraints: Acres: x+y ≤ 10 Quantity: tons of purple≥ 3 tons of orange 2x ≥ 12y Ex. Ordering Plants A store has a sale on pink and red flowers. There are two nurseries, East and West, that supply plants. East charges \$4/pink flower and \$6/red. West charges \$7/ pink and \$5/red. East supplies at most 200 of either colors, West supplies at most 400. You need at least 250 pink and at least 300 red. How many of each do you need to order to minimize the cost? 1. Variables: x= # of pink from East y=# of red from East z= # of pink from West q= # of red from West 2. Objective: minimize the cost C= 4x+6y+7z+5q 3. Constraints: # of total pink flowers: x+z250 # of total red flowers: y+q≥300 # of flowers from East: x+y ≤ 200 # of flowers from West: z+q ≤ 400 non-negativity: x ≥0 y ≥0 z ≥0 q ≥0 3.2- Graphing Linear Equations Steps: 1. Sketch the line using the x and y intercepts. The x-int is where y=0. Y-int is where x=0. 2. Decide if the line is solid or dashed. The line should be solid if it is ≤∨≥ . It should be dashed if it is ¿∨¿ . 3. Shade the appropriate side. Pick a test point on either side, making sure it’s not on the linear line. Shade the side where the plugged in variables keep the inequality correct. Feasible Region: region on the graph that has solutions to all the inequalities. Corner points: points that are at the corner of the feasible region. Ex. Hot drink cart- hot cocoa and lattes A latte costs \$3.75 and takes 3 minutes to produce. A hot chocolate costs \$2 and takes 2 minutes to produce. There are 450 cups and 960 minutes of production labor. How much of each should be made to maximize revenue? Constraints: x+y ≤ 450 cups 3x+2y ≤ 960 minutes x ≥ 0 y ≥ 0 non-negatives Intercepts: Inequality 1: x-int: x+0=450, x=450. (450,0) y-int: 0+y=450, y=450. (0,450) Inequality 2: 3x+2(0)=960 x= 320 (320,0) 3(0)+2y=960 y=480 (0, 480) Corner points: (0, 0), (320,0), (0,450) x+y=450, 3x+2y=960 intersect y=450-x 3x+2(450-x)=960 3x+900-2x=960 x=60 60+y=450 y=390 Corner point where they intersect: (60, 390)

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