Answer:
The correct option is;
C. 100
Step-by-step explanation:
Here we have
Maximum profit = 50X + 60Y
8X + 10Y ≤ 800 (labor hours)
X + Y ≤ 120 (total units demanded)
4X + 5Y ≤ 500 (raw materials)
All variables ≥ 0
Then we have
(i) 8X + 10Y ≤ 800
X = 0 or 100
Y = 80 or 0
(ii) X + Y ≤ 120
X = 0 or 120
Y = 120 or 0
(iii) 4X + 5Y ≤ 500
X = 0 or 125
Y = 100 or 0
from a chart of the above values we have in the most possible region, we have
Y intercept at X = 0 and Y = 80
Origin at X = 0, Y = 0
X intercept at X = 100 and Y = 0
Therefore, to maximize profit we have z(50X + 60Y)
At Y intercept (0, 80) = 4800
At origin = 0
At X intercept (100, 0) = 5000
Therefore, the number of units of regular model, X for maximum profit should be 100.
