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A manufacturer can sell x items at a price of Rs. \( \bigg( 5 - \large \frac{x}{100} \bigg) \) each. The cost price of x items is Rs. $ \bigg( \large\frac{x}{5}$$ + 500 \bigg) $. Find the number of items he should sell to earn the maximum profit.
cbse
class12
modelpaper
2009
sec-c
q29
math
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asked
Dec 29, 2012
by
thanvigandhi_1
edited
Jul 16, 2013
by
sreemathi.v
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A manufacturer can sell x items at a price of Rs. \( \bigg( 5-\large\frac{x}{100} \bigg) \) each. The cost price of x items is Rs. \( \bigg( \large\frac{x}{5} - \normalsize 500 \bigg) \). Find the number of items he should sell to earn maximum profit.
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A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 5760 to invest and has space for at most 20 items. A fan costs Rs. 360 and a sewing machine cost Rs. 240. He can sell a fan at a profit of Rs. 18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximise the profit? Translate the problem as an LPP and solve it graphically.
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Feb 10, 2013
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A dealer deals in two items only - item A and item B . He has $Rs. 50,000$ to inverst and a space to store at most 60 items. An item A cost $Rs. 2500$ and item B costs $Rs. 500$. A net profit to him on item A is $Rs. 500$ and on item B is $Rs. 150$. If he can sell all the items that he purchases , how should he inverst his amount to have maximum profit ? Formulate an LPP and solve is graphically .
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Mar 23, 2015
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