Question

B1 and B2 are two types of boats which are to be used to ferry 800 troops and 90 tons of equipment across a lake. Each B1 can carry 200 men and 15 tons of equipment while each B2 can carry 100 men and 15 tons of equipment. If each B1 costs Rs. 90 to operate and each B2 costs Rs. 44 to operate, find the number of each boat that should be used if the cost is to be minimum.

Answer 1

Let $x=$ the number of B1 boats, let $y=$ the number of B2 boats

$$x\geq0, y\geq0$$

The boats are used to ferry 800 troops and 90 tons of equipment across a lake

$$200x+100y\geq800$$

$$15x+15y\geq90$$

Therefore the mathematical formulation of the Linear Programming Problem is:

Minimize: $C = 90x+ 44y$

Subject to: $2x+y\geq8$

$x+y\geq6$

$x\geq0, y\geq0$

$AB:0\leq x\leq 2, y=8-2x$

$$C = 90x+ 44(8-2x)=2x+352$$ $$352\leq C \leq356$$

$BC:2\leq x\leq 6, y=6-x$

$$C = 90x+ 44(6-x)=46x+264$$

$$356\leq C \leq540$$

The cost has minimum with value of Rs.352, if we use only 8 B2 boats and don't use B1 boats.