Can anyone help me with this problem?An oil fired power station burns three grades of oil: grade A, grade B and grade C. The efficiency with which...

WHAT WE KNOW:

Just to see for element X, we know that we COULD get the (12 hrs/day times 200 tons of element X needed) 2400 tons of element X by buying 10,000 t(ons) of A (Grade A) oil (10,000t · .2t per t of A), plus 1000t of B (1000t · .1t per t of B), plus 1000t of C (1000t · .3t per t of C). This would give us 2000t from A, 100t from B, and 300t from C. It would cost us (10000t · £60/t) + (1000t · £80/t) + (1000t · £70/t) a total of (600,000 + 80,000. + 70,000) = £750,000. We do end up with exactly the 3600t of element Z, but we have 4500t of element Y, over by quite a bit from the 2100t needed.

CHECK FOR REASONABLENESS:

Would all A be best? All B? All C? We can see that B is the more expensive per ton and less productive than A for every element. We would only buy B if A is not available, hence we can eliminate B from consideration.

Let A = tons of Grade A oil, and C = tons of Grade C oil. From the example, we can derive:

Element X: 2400t = .2A + .3C, or 4800t = .4A + .6C, or 7200 = .6A + .9C
Element Y: 2100t = .4A + .2C, or 4200t = .8A + .4C

Element Z: 3600t = .3A + .4C, or 7200 = .6A + .8C

Graphing the above three lines, we could find either one, two, or three points of intersection. There could be none, or an infinite amount as well, which for a production problem such as this would not occur, since we have limits of 0≤A and C, and there is not a multiple of one equation.

Solving for Elements X & Y:

4800t = .4A + .6C

-(2100t = .4A + .2C)

2700t = + .4C, and dividing both sides by .4, we get 6750t needed of Grade C.

Substituting to find A, we get

2100 = .4A + .2·6750

2100 = .4A + 1350

750 = .4A, and 1875 = A

1st possible solution set: (A, C) = 1875t, 6750t

Cost is (1875t ∙ £60/t) + (6750t ∙ £70/t) = £585,000

Solving for Elements X & Z:

7200 = .6A + .9C

-(7200 = .6A + .8C)

0 = .1C, and C=0.

Substituting to find C, we get from Element X:

2400 = .2A + .3(0), or just 2400 = .2A, dividing both sides by .2, we get A = 12000.

2nd possible solution set: (A, C) = 12000t, 0t

Cost is (12000t ∙ £60/t) + (0t ∙ £70/t) = £720,000

Solving for Elements Y & Z:

4200 = .8A + .4C

-(3600 = .3A + .4C)

600 = .5A, and with dividing both sides by .5, we get A = 1200

Substituting to find C, we get from Element Z:

3600 = .3(1200) + .4C

3600 = 360 + .4C, subtract 360 from both sides,

3240 = .4C, and dividing both side by .4, we get C = 8100.

3rd possible solution set: (A, C) = 1200t, 8100t.

Cost is (1200t ∙ £60/t) + (8100t ∙ £70/t) = £574,200

The best solution is the last one. Remember, in a system of linear inequalities, the minimums or maximums occur at the intersections.