Problem 1:
Mrs. Burgos wants to buy at least 30 kilos of pork and beef for her restaurant business but can spend no more than Php 12, 000. A kilo of pork costs Php 180 and a kilo of beef costs Php 220. Give 4 possible amounts of pork and beef that Mrs. Burgos can buy.
Solution:
Let x be the kilos of pork
Let y be the kilos of beef.
x + y ≥ 30
180x + 220y ≤ 12000
if x = 0
x + y = 30
0 + y = 30
y = 30
y intercept (0, 30)
if y = 0
x + y = 30
x + 0 = 30
x = 30
x intercept (30, 0)
if x = 0
180x + 220y = 12000
180(0) + 220y = 12000
0 + 220y = 12000
y = 12000/220
y = 54.54545
y intercept (0, 54.54545)
if y = 0
180x + 220y = 12000
180x + 220(0) = 12000
180x + 0 = 12000
180x = 12000
x = 12000/180
x = 66.6667
x intercept (66.6667, 0)
plot these coordinates for the graph
y intercept (0, 30)
x intercept (30, 0)
y intercept (0, 54.54545)
x intercept (66.6667, 0)
The graph bounded by the shaded area is the solution to the inequality
This simply means that any point of ordered pairs within the shaded area is one of the possible solutions to both given equations.
Pick any point of ordered pairs inside the shaded area. The question is only asking for 4 possible solutions and these are:
(30, 20)
(30, 10)
(40, 10)
(50, 10)
Checking:
(30, 20)
x + y ≥ 30
30 + 20 ≥ 30
50 ≥ 30 ; true
180x + 220y ≤ 12000
180(30) + 220(20) ≤ 12000
5400 + 4400 ≤ 12000
9800 ≤ 12000 ; true
(30, 10)
x + y ≥ 30
30 + 10 ≥ 30
40 ≥ 30 ; true
180x + 220y ≤ 12000
180(30) + 220(10) ≤ 12000
5400 + 2200 ≤ 12000
7600 ≤ 12000 ; true
(40, 10)
x + y ≥ 30
40 + 10 ≥ 30
50 ≥ 30 ; true
180x + 220y ≤ 12000
180(40) + 220(10) ≤ 12000
7200 + 2200 ≤ 12000
9400 ≤ 12000 ; true
(50, 10)
x + y ≥ 30
50 + 10 ≥ 30
60 ≥ 30 ; true
180x + 220y ≤ 12000
180(50) + 220(10) ≤ 12000
9000 + 2200 ≤ 12000
11200 ≤ 12000 ; true
Problem 2:
Jane is buying squid balls and noodles for her friends. Each cup of noodles costs Php 15 while each stick of squid balls costs Php 10. She only has Php 70 but needs to buy at least 3 sticks of squid balls. Solve the system graphically.
Solution:
Let x be the noodles
Let y be the squid balls
15x + 10y ≥ 70
y ≥ 3
if x = 0
15x + 10y = 70
15(0) + 10y = 70
0 + 10y = 70
y = 70/10
y = 7 ;
y intercept (0,7)
if y = 0
15x + 10y = 70
15x + 10(0) = 70
15x + 0 = 70
15x = 70
x = 70/15
x intercept (70/15, 0)
Plot these coordinates
y intercept (0,7)
x intercept (70/15, 0)
if y = 3
15x + 10(3) = 70
15x + 30 = 70
15x = 70 - 30
15x = 40
x = 40/15
(40/15, 3) or (2.67, 3)
15(40/15) + 10(3) ≥ 70
40 + 30 ≥ 70
70 ≥ 70 ; true
(2.67, 3) ; Since we need a whole number value of cup noodle, instead of saying 2.67, we will be using 2 cups of noodles. This means that if she will buy at least 3 sticks of squid balls, she could still buy another 2 cups of noodles.
if y = 4
15x + 10(4) = 70
15x + 40 = 70
15x = 70 - 40
15x = 30
x = 2
(2, 4)
15(2) + 10(4) ≥ 70
30 + 40 ≥ 70
70 ≥ 70 ; true
This means that if she will buy at least 4 sticks of squid balls, she could still buy another 2 cups of noodles.
if y = 5
15x + 10(5) = 70
15x + 50 = 70
15x = 70 - 50
15x = 20
x = 20/15
(20/15, 5) or (1.33, 5)
15(20/15) + 10(5) ≥ 70
20 + 50 ≥ 70
70 ≥ 70 ; true
(1.33, 5) ; Since we need a whole number value of cup noodle, instead of saying 1.33, we will be using 1 cup of noodles. This means that if she will buy at least 5 sticks of squid balls, she could still buy another 1 cup of noodle.
if y = 6
15x + 10(6) = 70
15x + 60 = 70
15x = 70 - 60
15x = 10
x = 10/15
(10/15, 6) or (0.67, 6)
15(10/15) + 10(6) ≥ 70
10 + 60 ≥ 70
70 ≥ 70 ; true
(0.67, 6) ; Since we need a whole number value of cup noodle, instead of saying 0.67, we will be using 0 cup of noodles. This means that if she will buy at least 6 sticks of squid balls, she could no longer buy a cup of noodle.
if y = 7
15x + 10(7) = 70
15x + 70 = 70
15x = 70 - 70
15x = 0
x = 0
(0, 7)
15(0) + 10(7) ≥ 70
0 + 70 ≥ 70
70 ≥ 70 ; true
This means that if she will buy at least 7 sticks of squid balls, she could no longer buy a cup of noodle.
#CarryOnLearning
