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.
Graph C is the answer.
2x – y ≥ 3x + 2y < 4
=y is less than or equal to 2x-3
y is less than -x/2+2
Answer:
1 < x
Step-by-step explanation:
6x - 12 < 8x- 14
Subtract 6x from each side
6x -6x- 12 < 8x-6x- 14
-12 < 2x -14
Add 14 to each side
-12 +14 < 2x
2 < 2x
Divide each side 2
2/2 < 2x/2
1 < x
15 performances can be accommodated
<em><u>Solution:</u></em>
Given that,
Each performer for the talent show has 7 minutes for his or her performance
The introduction for the talent show is 25 minutes long
The entire show including the introduction, will last 130 minutes
Thus the remaining time for performances is:
Performance time = 130 minutes - 25 minutes
Performance time = 105 minutes
To find, the number of performances, divide the 105 by 7 minutes

Thus 15 performances can be accommodated