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kaplan revision kit question, 15- B Chemicals
In this question there are two products super and regular petrol and two constraints light and heavy crude.
in this question there is another constraint of market condition which suggests that atleast two thirds of the refined petrol should be super petrol.
the total amount of petrol, regular + super is 11000 gallons. in this question i am unable to understand what would be the equation and how would it reflect in the graph, in the identification of feasible region.this is not tough one.
let x gallons be super petrol to produce
let y gallons be regular petrol to produce.0.7 x+ 0.5 y <= 5000
0.5 x+ 0.7 y<= 6000 2/3(x+y) <= x
i.e. 2(x+y)<=3x
i.e 2x +2y<=3x
2y<=xyes, this is what i came across in answer but just i am not able to figure out how it will effect the feasible area in the graph what would we plot in graph for this constraint ? and can we do like taking 2/3 of 11000 gallons and plot a line at 7333 on x axis being for super petrol and 1/3 of it 3667 on y axis ?
@sids, yes i have gone through but my answer page where the graph is plotted is misprinted so here i am. And, this the only such adjustment i found in the kit where there is a constraint where the product must be more or equal to the constraint so some doubts. Any problem ?
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