We are asked to find the least amount of time that machine B has
to work. For this condition, machine A has to operate for max hours i.e. 8
100,000 = 8 * 9000 + x * 7000
x = 4 hours
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This one seems simple
For Machine B to run for the least possible time, the maximum work should be delegated to Machine A
Thus assume that Machine A works for 8 hours (maximum permissble time per machine)
In 8 hours A will produce 9000 * 8 = 72,000 pencils
Pencils left to be manufactured = 100,000 - 72000 = 28,000
In 1 Hour B can make 7000 pencils
Therefore B will make 28,000 pencils in 28,000/7000 = 4 Hours
For Machine B to run for the least possible time, the maximum work should be delegated to Machine A
Thus assume that Machine A works for 8 hours (maximum permissble time per machine)
In 8 hours A will produce 9000 * 8 = 72,000 pencils
Pencils left to be manufactured = 100,000 - 72000 = 28,000
In 1 Hour B can make 7000 pencils
Therefore B will make 28,000 pencils in 28,000/7000 = 4 Hours
if machine B's operational time is to be minimized, we must maximize the time for machine A to operate. Since each machine can operate for at most 8 hours, we can let the machine A operate for 8 hours.
since the rate of machine A is 9000 pencils per hour, machine A produces 8 * 9000 = 72000 pencils, and thus 28000 pencils are left to be produced.
Thus, it will take machine B
28000/7000 = 4 hours to produce the remaining pencils
since the rate of machine A is 9000 pencils per hour, machine A produces 8 * 9000 = 72000 pencils, and thus 28000 pencils are left to be produced.
Thus, it will take machine B
28000/7000 = 4 hours to produce the remaining pencils
