Answer:
The weekly profit function P(x,y) = 72x - 0.01x² -0.005xy + 38y - 0.005xy -0.015y² - 10000
3.5 oz bars = 3560
5 oz bars = 80
The maximum weekly profit = 119680
Step-by-step explanation:
Given that:
the demand functions are p and q
the variable cost of x = 8x and the variable cost of y = 12y
Since the fixed cost attributed to the division is exist 10,000/week
∴ the total cost C = 8x + 12y + 10000
R(x,y) = p(x) + q(y)
where ;
xp(x) = 80x -0.01x² -0.005y
yqz(y) = 50 -0.005x -0.015y²
R(x,y) = 80x -0.01x² -0.005xy + 50y -0.005xy -0.015y²
The weekly profit function P(x,y) = R(x,y) - C
P(x,y) = 80x -0.01x² -0.005xy + 50 -0.005xy -0.015y² - ( 8x + 12y + 10000)
P(x,y) = 80x -0.01x² -0.005xy + 50 - 0.005xy -0.015y² - 8x -12y - 10000
P(x,y) = 72x - 0.01x² -0.005xy + 38y - 0.005xy -0.015y² - 10000
(b) How many of the 3.5-oz size bars and how many of the 5-oz bath size bars should the division produce per week to maximize its profit?
profit will be maximized if:
[tex]\dfrac{\partial P}{\partial x} =0 \ and \ \dfrac{\partial P}{\partial y} =0[/tex]
given that:
[tex]\dfrac{\partial^2 P}{\partial x^2} < 0 \ and \ D >0[/tex]
where;
D = Px Py -Pxy²
Now;
[tex]\dfrac{\partial P}{\partial x} =(-0.02 x -0.01 y +72)[/tex]
[tex]\dfrac{\partial P}{\partial y} =(-0.03 x -0.01 y +38)[/tex]
[tex]\dfrac{\partial P}{\partial x} =0 \to 2x+y = 7200[/tex]
[tex]\dfrac{\partial P}{\partial y} =0 \to x+3y = 3800[/tex]
2x + y = 7200
2x + 6y = 7600
5y = 400
y = 400/5
y = 80
From
x + 3y = 3800
x + 3(80) = 3800
x = 3800 - 240
x = 3560
∴ 3.5 oz bars = 3560
5 oz bars = 80
What is the maximum weekly profit?
The maximum weekly profit = 72x - 0.01x² -0.005xy + 38y - 0.005xy -0.015y² - 10000
The maximum weekly profit = 72(3560) - 0.01(3560)² -0.005(3560)(80) + 38(80) - 0.005(3560)(80) -0.015(80)² - 10000
The maximum weekly profit = 256320 - 126736 - 1424 + 3040 - 1424 -96 - 10000
The maximum weekly profit = 119680
