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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 -

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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

MCQ

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.

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