Answer:
The maximum profit is when they make 10 units of A and 2 units of B.
Step-by-step explanation:
Let x is units of milk
Let y units of cacao
Given that :
The company's production plant has a total of 22 units of milk and 46 units of cacao available.
2x + y ≤ 22 (2 unit of milk for each of A and 1 for B; 22 units available)
4x + 3y ≤46 (4 unit of milk for each of A and 3 for B; 46 units available
Graph the constraint equations and find the point of intersection to determine the feasibility region.
The intersection point (algebraically, or from the graph) is (10, 2)
The objective function for the problem is the total profit, which is $6.2 per unit for A and $4.2 per unit for B: 6.2x + 4.2y.
Hence, we substitute (10, 2) into the above function:
6.2*10 + 4.2*2 = 70.4
The maximum profit is when they make 10 units of A and 2 units of B.
The linear programming model is,
[tex]\rm 2x +y \leq 22[/tex] and [tex]\rm 4x +3y \leq 46[/tex]
The manufacturing quantity for 10 units of A and 2 units of B in order to maximize profit is 70.4.
Given that,
A chocolate manufacturing company produces two types of chocolate: A and B.
Ingredients required for manufacturing the products include milk and cacao only.
Each unit of type A chocolate requires 2 units of milk and 4 units of cacao.
Each unit of type B chocolate requires 1 unit of milk and 3 units of cacao.
The company's production plant has a total of 22 units of milk and 46 units of cacao available.
On each sale, the company makes a profit of $6.20 for every unit of chocolate of type A and $4.20 for every unit of type B.
We have to determine,
Develop a linear programming model to determine the manufacturing quantity for each type in order to maximize profit.
According to the question,
Let, x be the unit of milk used,
And y be the unit of cacao used,
Ingredients required for manufacturing the products include milk and cacao only.
Each unit of type A chocolate requires 2 units of milk and 1 unit of milk for type B chocolate.
The company's production plant has a total of 22 units of milk.
Therefore,
Linear programming is,
[tex]\rm 2x +y \leq 22[/tex]
And each unit of type A chocolate requires 4 units of cocoa and 3 unit of cocoa for type B chocolate.
The company's production plant has a total of 46 units of cocoa.
Therefore,
Linear programming is,
[tex]\rm 4x +3y \leq 46[/tex]
Then,
The manufacturing quantity for each type in order to maximize profit is,
[tex]\rm P= 6.2x + 4.2y[/tex]
From the graph constraint and find the point of intersection to determine the feasibility region.
The point of intersection to determine the feasibility region is (10, 2) shown in the attached graph.
Therefore,
The manufacturing quantity for each type in order to maximize profit is,
[tex]\rm P= 6.2x + 4.2y[/tex]
The feasibility region is at point (10, 2)
[tex]\rm P= 6.2x + 4.2y\\\\\rm P= 6.2(10)+ 4.2(2)\\\\P = 62 + 8.4\\\\P = 70.4[/tex]
Hence, The manufacturing quantity for 10 units of A and 2 units of B in order to maximize profit is 70.4.
For more details refer to the link given below.
https://brainly.com/question/17444642
