Let number of kgs. of fertilizer F1 = x
And number of kgs. of fertilizer F2 = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 6x +5y , subject to the constraints : 10/100 x + 5/100y ≥ 14 and 6/100x + 10/100y ≥ 14, i.e. 2 x + y ≥ 280 and 3x + 5y ≥ 700, x,y ≥ 0.,
Corner points | Z =6x +5 y |
A ( 0 , 280 ) | 1400 |
D(700/3,0 ) | 1400 |
B(100,80) | 1000………….(Min.) |
Corner points Z =6x +5 y A ( 0 , 280 ) 1400 D(700/3,0 ) 1400 B(100,80) 1000………….(Min.) Here Z = 1000 is minimum.
i.e. 100 kg of fertilizer F1 and 80 kg of fertilizer F2; Minimum cost = Rs 1000.
Let number of bags of cattle feed of brand P = x
And number of bags of cattle feed of brand Q = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 250x +200y , subject to the constraints : 3 x + 1.5y ≥ 80, 2.5x + 11.25y ≥ 45, 2x + 3y ≥ 24 , x,y ≥ 0.,
Corner points | Z =250x +200 y |
C(0 , 12 ) | 2400
|
B (18,0) | 4500 |
D(3,6 ) | 1950…………………(Min.) |
A(9,2) | 2650 |
Here Z = 1950 is minimum.
i.e. 3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950 .
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food | Vitamin A | Vitamin B | Vitamin C |
X | 1 | 2 | 3 |
Y | 2 | 2 | 1 |
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?
Let number of kgs of food of brand X = x
And number of kgs. of food of brand Y = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 16x +20y , subject to the constraints : x + 2y ≥ 10,2x + 2y ≥12,3x + y ≥8,x,y ≥ 0.,
Corner points | Z =16x +20 y |
C(10 , 0 ) | 160
|
B (0,8) | 160 |
D(1,5 ) | 116 |
A(2,4) | 112……………..(Min.) |
Here Z = 112 is minimum.
i.e. Least cost of the mixture is Rs 112 (2 kg of Food X and 4 kg of food Y).
Let number of executive class tickets = x
And number of economy class tickets = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 1000x +600y , subject to the constraints : x + y ≤ 200, 4x - y ≤ 0, x ≥ 20,x,y ≥ 0.,
Corner points | Z =1000x +600 y |
C(20 ,80 ) | 68000
|
B (40,160) | 136000 |
D(20,180 ) | 12800 |
Here Z = 136000 is maximum.
i.e. 40 tickets of executive class and 160 tickets of economy class; Maximum profit = Rs 136000 .
Here the objective function is given by : F = 4x +6y .
Corner points | Z = 4x +6 y |
(0, 2 ) | 12………………..(Min.) |
(3,0) | 12………………….(Min.) |
(6,0 ) | 24 |
(6 , 8 ) | 72 |
(0 , 5 ) | 30 |
Maximum of F – Minimum of F = 72 – 12 = 30 .
Here , maximize Z = 3x+4y ,
Corner points | Z = 3x + 4 y |
C( 0 ,38 ) | 132 |
B ( 52 ,0) | 156 |
D(44 , 16) | 196 ………………….(Max.) |
Hence the maximum value is 196
Feasible region (shaded) for a LPP is shown in Figure. Maximize Z = 5x + 7y.
Corner points | Z = 5x +7 y |
O(0, 0 ) | 0
|
B (3,4) | 43…………………..(Max.) |
A(7,0 ) | 35 |
C(0,2) | 14 |
Hence the maximum value is 43