5(2x-3)-(x+8)=13
10x-15-x-8=13
9x-23=13
9x=36
x=4
The answer should be 4.3 (the line is ontop of the 3)
The line is on the 3
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.
Answer:
do you need all work shown for this problem
Answer:costC(p)= 1400+20p
R(p)= 4p^2+200p
Profit= 4p^2+ 180p-1400
Step-by-step explanation:
q=4p+200
Student council charge $400 per week.
Revenue= price × quantity.
Relationship between revenue and profit= Revenue-total cost.
Total cost= cost of facilities +(cost of one shirt)× number of T-shirts sold per week.
C(p)= 400 + 5(q)
Put q=4p+200
C(p)= 400 + 5 (4p +200)
C(p)= 400+ 20p + 1000
C(p)= 1400+ 20p
Relationship between revenue and price=Revenue= price × quantity
R(p)= p×q
Put q=4p +200
R(p)= p ×(4p +200)
R(p)= 4p^2+ 200p
Profit= Revenue - total cos
Profit=R(p) -C(p)
Profit= (4p^2 + 200p) - ( 20p +1400)
Profit= 4p^2 + 180p -1400