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A manufacturer produces nuts and bolts. It takes 1 hours of work on machine. A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hours -

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प्रश्न

A manufacturer produces nuts and bolts. It takes 1 hours of work on machine. A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hours on machine B to produce a packages of bolts. He earns a profit of Rs. 17.50 per packages on nuts and Rs. 7.00 per packages on bolts. How many packages of each should be produced each day so as to maximise his profit if he operates his machine for at the most 12 hours a day?

विकल्प

  • 2 nuts 2 bolts

  • 3 nuts 3 bolts

  • 4 nuts 4 bolts

  • 5 nuts 5 bolts

MCQ

उत्तर

3 nuts 3 bolts

Explanation:

We have if x nuts and y botts are produced. the following information:

Item Number Machine
A
Machine
B
Profit
Nuts x 1 hours 3 hours 17.50
Bolts y 3 hours 1 hours 7.00
Max
Time
available
  12 hours    

Machine A is used for x × 1 + y × 3 hours

Maximum time available = 12 hours

∴ x + 3y ≤ 12

Machine B is used for 3 × x + 1 × y hours

Maximum available time is 12 hours.

⇒ 3x + y ≤ 12

Profit function z = `(17.50) xx x + (7.00)y`

Thus the objective function is z = 17. 5x + 7y constraints are x + 3y ≤ 12, 3x + y ≤ 12 and x, y ≥ 0

(i) The line x + 3y = 12, passes through A(12, 0), B(0, 4) putting x = 0, y = 0 in x + 3y ≤ 12 we get 0 ≤ 12 is true.

(ii) The line 3x + y = 12 passes through C(4, 0) and D(0, 12) putting x = 0, y = 0 in 3x + y ≤ 12, we get 0 ≤ 12 which is true 3x + y ≤ 12 lies on and below CD.

(iii) x ≥ 0 is the region which lies on and to the right of y-axis·

(iv) y ≥ 0 is the region which lies on and above the x-axis.

(v) The feasible region is shaded as BPCO. The intersection of the lines is marked by the letter 'P.'

AB : x + 3y = 12  ......(i)

CD : 3x + y = 12 ......(ii)

Multiplying equation (i) by 3 and subtracting (ii) from it.

8y = 36 – 12 = 24, ∴ y = 3

From (i) x = 12 – 3y = 12 – 9 = 3

∴ The point  P is (3, 3)

Z = 17.5x + 7y

At B(0, 4) Z = 0 + 7 × 4 = 28
At C(3, 3) Z = 17.5 × 3 + 7 × 3 = 52.5 + 21 = 73.5
At C(4, 0) Z = 17.5 × 4 × 0 = 70
At O(0, 0) Z = 0

When three packages of nuts and bolts are produced, the maximum profit is Rs.73.50

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