Let the number of units of shoe A be x and shoe B be y, time taken by shoe A is hours & time taken by shoe B is 1 hour company can produce 650 pairs per day
∴ 3x + y ≤ 650
Similarly maximum of 150 pairs of shoe A can be produced i.e x ≤ 150 and 400 of shoe B can be sold i.e. y ≤ 400
Maximize z = 400x + 105y subject to the constraints
3x + y ≤ 650
x ≤ 150
y ≤ 400
x ≥ 0,y ≥ 0