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NCERT Exemplar · Q19

Q.Refer to Exercise 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

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For the sweaters LPP of Exercise 14, profit Z=200x+120yZ=200x+120y is greatest at (150,150)(150,150), giving ₹48000.

The referenced problem (Exercise 14)

A company makes Type A sweaters (cost ₹360) and Type B (cost ₹120). It can make at most 300300 sweaters and spend at most ₹72000 a day, and the number of Type B cannot exceed the number of Type A by more than 100100. Profit is ₹200 on A and ₹120 on B. Let xx = Type A, yy = Type B.

Maximise Z=200x+120y\text{Maximise } Z=200x+120y

360x+120y≤72000,x+y≤300,y−x≤100,x,y≥0,360x+120y\le72000,\quad x+y\le300,\quad y-x\le100,\quad x,y\ge0,

which simplify (divide the first by 120120) to

3x+y≤600,x+y≤300,y−x≤100.3x+y\le600,\qquad x+y\le300,\qquad y-x\le100.

Step 1 — Corner points

  • (0,0)(0,0).
  • 3x+y=6003x+y=600 with y=0⇒(200,0)y=0\Rightarrow(200,0) (check x+y=200≤300x+y=200\le300).
  • 3x+y=6003x+y=600 and x+y=300x+y=300: subtract ⇒2x=300⇒x=150, y=150⇒(150,150)\Rightarrow2x=300\Rightarrow x=150,\,y=150\Rightarrow(150,150) (check y−x=0≤100y-x=0\le100).
  • x+y=300x+y=300 and y−x=100y-x=100: add ⇒2y=400⇒y=200, x=100⇒(100,200)\Rightarrow2y=400\Rightarrow y=200,\,x=100\Rightarrow(100,200) (check 3x+y=500≤6003x+y=500\le600). …

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