A company produces two different products X and Y makes a profit of ₹30 and ₹40 per unit. Production process has a capacity of 30000 hours. It takes 3 hours to produce 1 unit of X and 1 hour for 1 unit of Y. Maximum unit which can be sold 12000 for X and 8000 for Y.
Formulate the problem and solve it by graphical method to get maximum profits.
Answers
Answer:
The company should produce 2 units of type A and 3 units of type B to maximize the profit and the maximum profit will be Rs. 230
Step-by-step explanation:
1
Secondary School Math 13 points
A company produces two types of goods a and b, that require gold and silver.Each unit of type a requires 3g of silver and1g of gold while that of b requires 1g of silver and 2g of gold.The company can use atmost 9g of silver and 8g of gold.If each unit of type a bring a profit of rupees 40 and that of type b rupees 50,find the number of units of each type that the company should produce to maximize the profit.Formulate and solve graphically the lpp and find maximum profit in . Com
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The company should produce 2 units of type A and 3 units of type B to maximize the profit and the maximum profit will be Rs. 230
Explanation
Suppose, the number of units of type A is x and the number of units of type B is y
Each unit of type A requires 3 g of silver and 1 g of gold while that of B requires 1 g of silver and 2 g of gold.
So, the total amount of silver in two types =(3x+y) g
and the total amount of gold in two types =(x+2y) g.
The company can use at most 9 g of silver and 8 g of gold.
constraints are......
Now, each unit of type A bring a profit of Rs. 40 and that of type B Rs. 50. So, the profit function will be:
If we graph the constraints, then the vertices of the common shaded region
will be: (0,0), (3,0), (2,3) and (0,4) (Please refer to the below attached image for the graph)
For (0, 0) ⇒ P= 40(0)+50(0)=0
For (3, 0) ⇒ P= 40(3)+50(0)=120
For (2, 3) ⇒ P= 40(2)+50(3)=80+150=230
For (0, 4) ⇒ P= 40(0)+50(4)=200
So, the profit will be maximum when x=2 and y=3