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# Grade 10 Math: Linear Programming Assignment

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## Matheesan Thirukumaran

on 6 March 2013

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#### Transcript of Grade 10 Math: Linear Programming Assignment

Linear Programming Assignment By: Matheesan T. 10F2 References Step 1:
Defining the Variables Let x represent the number of dining tables.
Let y represent the number of patio tables.
Let P represent the amount of profit. Step 2:
Create the Objective Function Maximize Profit:
P=300x + 250y Dining Tables Patio Tables Total Profit (\$) 300x 250y 300x + 250y Cutting Time Assembly Time Sanding Time Quality Control Panting Time 30 25 300 40 30 300 20 20 180 10 15 120 15 10 120 Step 3:
Constraints 1.300≥40x+30y (Cutting Time)
2.300≥30x+25y (Assembly Time)
3.180≥20x+20y (Sanding Time)
4.120≥10x+15y (Quality Control)
5.120≥15x+10y (Painting Time)
6. x≥0 (Non-negativity)
7. y≥0 (Non-negativity) Step 4:
Finding the x&y intercepts Equation # X-Intercept Y-Intercept 1. 2. 3. 4. 5. 7.5 10 10 12 9 9 12 8 8 12 1. y≤-4/3x+10
2.y≤-6/5x+12
3.y≤-x+9
4.y≤-2/3x+8
5.y≤-3/2x+12 Step 5:
Changing Constraints into Y-intercept form Since the y value is less than or equal, it means that I will have to shade below the graphed lines to find the feasible region. Step 6:
Graphing and Shading the Feasible Region Step 7:
Finding the Corner Points A (0,8) Equation 4
B (3,6) Intersection of Equation 1,3,4
C (7.5,0) Equation 1
O (0,0) Origin 1. 300≥40x+30y
3. (180≥20x+20y)2

300=40x+30y
360=40x+40y (subtraction)
-----------------
-60=-10y
-60/-10=-10y/-10
6=y

40x+30(6)=300
40x+180=300
40x=120
40x/40=120/40
x=3 1. 300≥40x+30y
4. (120≥10x+15y)4

300=40x+30y
480=40+60y (subtraction)
-----------------
-180=-30y
-180/-30=-30y/-30
6=y

10x+15(6)=120
10x+90=120
10x=30
10x/10=30/10
x=3 Finding B: Step 8:
Finding the Optimal Values Objective Function-->Maximize Profit:
P=300x + 250y A(0,8)

P=300(0)+250(8)
=\$2000 B(3,6)

P=300(3)+250(6)
=900+1500
= \$2400 C (7.5,0)

P=300(7.5)+250(0)
=\$2250 O (0,0)

P=300(0)+250(0)
=\$0 Step 9:
Stating the Optimal Values The Problem: I run a small table manufacturing company. Our company manufacturers two types of tables, dining and patio. We make \$300 profit from each dining table we sell and \$250 profit for each patio table we sell.

Each type of table goes through the same five steps; cutting, assembly, sanding, quality control, and painting. It takes 40 minutes to cut out all the pieces for a dining table and 30 minutes for a patio table. The CNC router machine can only be operated for a maximum of 5 hours per day. After the pieces are cut, they are assembled by hand. It takes my workers 30 minutes to assemble a dining table and 25 minutes to assemble a patio table. The maximum amount of time on assembly is 5 hours in a day. Tables are then sanded down to become smooth. It takes 20 minutes for both tables and the maximum time spent sanding in a day is 2hours and 30 minutes. Both tables then go through quality control for 15 minutes. The maximum time for this is 2 hours per day. Lastly, the tables are painted accorded to the request. It takes 15 minutes to paint a dining table and 10 minutes to paint a patio table. Painting has a limit of 2 hours.

What is the optimal amount of each table to gain the largest profit per day assuming that all the tables made are sold? The optimal number of dining tables and patio tables to make the highest profit of \$2400 would be 3 and 6 respectively. The Solution: In order for the company to obtain its maximum profit of \$2400 per day it must manufacture 3 dining tables and 6 patio tables. Finding the Optimal Values: The information and values used in the problems are from my cousin's company. Links Used: