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Question
One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake require 100 g of flour and 50 kg fat. Find the mamximum number of cake which can be made from 5 kg of flour and l kg of fat assuming that there is no shortage of the other ingradients used in making the cakes.
Options
30
40
45
50
Solution
30
Explanation:
Let x cakes of I kind and y cakes of II kind are being made.
Now we have
Kind of cake | No. of cakes | Flour | Fat |
I | x | 200 g | 25 g |
II | y | 100 g | 50 g |
Total | x+y | 500 g | 1000 g |
Now objective function is z = x + y
Maximum quantity of flour used = 5000 g
⇒ 200 x + 100 y ≤ 5000 g
or 2x + y ≤ 50 ......(i)
Maximum quantity of fat used = 1000 g
25x + 50y ≤ 1000
x + 2y ≤ 40 ......(ii)
x, y ≥ 0 ........(iii)
(i) The line 2x + y = 50 passes through the point A(25, 0), B(0, 50)
Putting x = 0, y = 0 in 2x + y ≤ 50, we get (0, 50), which is true
⇒ 2x + y ≤ 50 represent the region or and below the line AB
(ii) The line x + 2y = 40 passes through C(40, 0), D(0, 20) putting x = 0, y = 0 in x + 2y ≤ 40
We get 0 ≤ 40, which is true
⇒ x + 2y
⇒ 40 represent the region on and below CD
(iii) x ≥ 0 is the region on and on the right of y-axis.
(iv) y ≥ 0 is the region on and above x-axis.
Feasible region is OAPD which is a bounded region shown by shaded area.
The line AB : 2x + y = 50 and CD : x + 2y = 40 intersect at P.
Solving there equations we get x = 20 and y = 10.
∴ The point P is (20, 10)
At A(25, 0) | z = x + y = 25 + 0 = 25 |
At P(20, 10) | x = 20 + 10 = 30 |
At (0, 20) | z = 0 + 20 = 20 |
At (0, 0) | z = 0 |
Hence maximum value of z is 30 at P(20, 10) i.e., I kind of cakes are 20 and II kind of cakes 10 will maximum z.