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+1 vote
9.4k views
in Linear Programming by (65.5k points)
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A company owned by Navya manufactures two types of cloth, using three different colours of wool. One yard length of type A cloth required 4 oz(once) of red wool, 5 oz of green wool, 3 oz (ounce) of yellow wool. One yard length of type B cloth requires 5 oz red wool, 2 oz of green wool and 8 oz of yellow wool. The wool available for manufacture is 1000 oz of red wool 1000 oz of green wool and 1200 oz of yellow wool. The manufactures can make a profit of Rs 5 on one yard of type A cloth and Rs 3 on one yard of type B cloth. Find the best combination of the quantities of type A and Type B cloth which gives him maximum profit by solving the L.P.P. by graphical method.

2 Answers

+1 vote
by (44.0k points)
selected by
 
Best answer

max z = 5x + 3y

Constraint

4x + 5y ≤ 1000

5x + 2y ≤ 1000

3x + 8y ≤ 1200

x,y ≥ 0

intersection point of line (1) and (2)

5 x equation (2) - 2 equation(1) ⇒ 25x + 16y - 8x - 16y = 5000 - 2000

⇒ 17x = 3000 ⇒  x = \(\frac{3000}{17}\)

5\(\big(\frac{3000}{17}\big)\) + 2y = 1000

⇒ 2y = 1000 - \(\frac{15000}{17}\) = \(\frac{2000}{17}\) 

⇒ y = \(\frac{1000}{17}\)

\(\big(\frac{3000}{17},\frac{1000}{17}\big)\)

intersection point of line (2) and (3)

4 x equation (2) - equation (3)

⇒ 20x + 8y - 3x - 8y = 4000 - 1200

⇒ 17x = 2800 ⇒ x = \(\frac{2800}{17}\)

2y = 1000 - 5\(\big(\frac{2800}{17}\big)\) = \(\frac{17000-14000}{17}\) = \(\frac{3000}{17}\) 

y = \(\frac{1500}{17}\)

\(\big(\frac{2800}{17},\frac{1500}{17}\big)\)

intersection point of (1) and (3)

4 x equation(3) - 3 x equation(1) ⇒ 12x + 32y - 12x - 15y = 4800 - 3000

⇒ 17y = 1800

⇒ y = \(\frac{1800}{17}\)

4x = 1000 - 5y = 1000 - 5\(\big(\frac{1800}{17}\big)\) = \(\frac{17000-9000}{17}\)

x = \(\frac{2000}{17}\)

\(\big(\frac{2000}{17},\frac{1800}{17}\big)\)

Clearly by graph corner points are

(0,0), (0,150), (\(\big(\frac{2000}{17},\frac{1800}{17}\big)\),\(\big(\frac{3000}{17},\frac{1000}{17}\big)\) and (200,0)

Points z = 5x + 3y
(0,0) 0
(0,150) 450
\(\big(\frac{2000}{17},\frac{1800}{17}\big)\) \(\frac{15400}{17}\) = 905.88
\(\big(\frac{3000}{17},\frac{1000}{17}\big)\) \(\frac{18000}{17}\) = 1058.82 (maximum)
(200,0) 1000

Therefore, maximum profit = 1058.82 at point \(\big(\frac{3000}{17},\frac{1000}{17}\big)\).

+4 votes
by (65.2k points)

Maximize Z = 5x + 3y 

subject to 

4x + 5y ≤ 1,000 

5x + 2y ≤ 1,000 

3x + 8y ≤ 1,200 

x, y ≥ 0 

Plot the lines on the graph & solution z is maximum at (0, 200) & zmax= 6,000.

by (10 points)
Pls give me all solution

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