
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and a half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs. 100 and that on a bracelet is Rs. 300. Formulate an L.P.P. for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.
Answer
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Hint: Let x be the number of necklaces manufactured and y be the number of bracelets manufactured. Using the condition: the total number of necklaces and bracelets that it can handle per day is at most 24, form one constraint. Using maximum number of hours form another constraint. Then using the condition that at least one of each must be produced, form two more constraints. These will be the four constraints. Next, let the profits be z. Use the condition that the profit on a necklace is Rs. 100 and that on a bracelet is Rs. 300 to form an equation which needs to be maximised. This is the L.P.P.
Complete step-by-step answer:
In this question, we are given that a small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and a half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs. 100 and that on a bracelet is Rs. 300.
We need to formulate an L.P.P. for finding how many of each should be produced daily to maximize the profit.
Let x be the number of necklaces manufactured and y be the number of bracelets manufactured. The total number of necklaces and bracelets it can handle is at most 24, so, according to the question,
$x+y\le 24$
The necklaces take $x$ hours to manufacture and bracelets take $\dfrac{y}{2}$ hours to manufacture and the maximum time available is for 16 hours. So,
\[x+\dfrac{y}{2}\le 16\]
The profit on one necklace is given as Rs. 100 and the profit on one bracelet is given as Rs. 300.
Let the profit be z. So,
$z=100x+300y$
Therefore, we need to maximize
$z=100x+300y$
Subject to the constraints
$x+y\le 24$ and
\[x+\dfrac{y}{2}\le 16\]
$x\ge 1$ and $y\ge 1$
max $z=100x+300y$ is the required L.P.P.
Note: In this question, it is very important to note that the question asks us to formulate an L.P.P. only to find how many of each should be produced daily to maximize the profit. We do not actually have to find the values of x and y or the maximum profit. Some students may not understand this and solve it to get the values which will waste their time.
Complete step-by-step answer:
In this question, we are given that a small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and a half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs. 100 and that on a bracelet is Rs. 300.
We need to formulate an L.P.P. for finding how many of each should be produced daily to maximize the profit.
Let x be the number of necklaces manufactured and y be the number of bracelets manufactured. The total number of necklaces and bracelets it can handle is at most 24, so, according to the question,
$x+y\le 24$
The necklaces take $x$ hours to manufacture and bracelets take $\dfrac{y}{2}$ hours to manufacture and the maximum time available is for 16 hours. So,
\[x+\dfrac{y}{2}\le 16\]
The profit on one necklace is given as Rs. 100 and the profit on one bracelet is given as Rs. 300.
Let the profit be z. So,
$z=100x+300y$
Therefore, we need to maximize
$z=100x+300y$
Subject to the constraints
$x+y\le 24$ and
\[x+\dfrac{y}{2}\le 16\]
$x\ge 1$ and $y\ge 1$
max $z=100x+300y$ is the required L.P.P.
Note: In this question, it is very important to note that the question asks us to formulate an L.P.P. only to find how many of each should be produced daily to maximize the profit. We do not actually have to find the values of x and y or the maximum profit. Some students may not understand this and solve it to get the values which will waste their time.
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