Question

Convert the inequalities into equations, and then use the substitution method to find four possible vertices in the form (x, y, z).

Answer 1

For 2x + 5y < 100, x + y < 30. Vertices : (50,0) , (0,20) , (35,0) , (0,35) , (25,0)

Explanation:

Considering inequality equations

i) 2x + 5y < 100 ;

ii) x + y < 35

Concerting them to equality conditions

i) 2x + 5y =100.

Intercepts : If y = 0, x = 50. So, (50,0). If x = 0, y = 20. So, (0,20)

ii) x + y = 35. Intercepts :

Intercepts : If y = 0, x = 35. So, (35,0). If x = 0, y = 35 So, (0,35)

• Solving them as per equality (with substitution)

From ii), x = 35 - y . Putting the value in i)

2 (35 - y) + 5y = 100 → 70 - 2y + 5y = 100 → 70 + 3y = 100

3y = 100 - 70 → 3y = 30 → y = 30 / 3 = 10

Putting y = 10 in ii) : x = 35 - 10 = 25

• As per the 'less than' inequalities : the inner 'towards origin' area from both lines is in common shaded area.

Henceworth, vertices are : (50,0) , (0,20) , (35,0) , (0,35) & intersection (25,0)

Answer 2

Correct answers for edmentum & plato:

Explanation:

Remember:

x+y+z=60

x=5

y=30

z=15

1:

x=5y=30

5+30+z=60

35+z=60

-35+z=60(-25)

z=25

(5,30,25)

2:

x=5z=15

5+y+15=60

20+y=60

-20+y=60-(20)

y=40

(5,40,15)

3:

y=30z=15

x+30+15=60

x+45=60

x+-45=60-(45)

x=15

(15,30,15)

4:

(5,30,15)

5+30+15=60

50=60

-50=60-(50)

10

(5,30,15) Doesn't work