Answer:
For 2x + 5y <u>< </u>100, x + y <u><</u> 30. Vertices : (50,0) , (0,20) , (35,0) , (0,35) , (25,0)
Step-by-step explanation:
Considering inequality equations
i) 2x + 5y <u>< </u>100 ;
ii) x + y <u><</u> 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, <u>vertices </u>are : (50,0) , (0,20) , (35,0) , (0,35) & intersection (25,0)
