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

Answer:

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

Step-by-step 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)