Mathematics, 27.08.2020 01:01 chantelljenkins2

Johnson's Household Products has a division that produces two sizes of bar soap. The demand equations that relate the prices p and q (in dollars per hundred bars), to the quantities demanded, x and y (in units of a hundred), of the 3.5-oz size bar soap and the 5-oz bath size bar soap are given by p = 80 - 0.01x - 0.005y and q = 50 - 0.005x - 0.015y. The fixed cost attributed to the division is exist10,000/week, and the cost for producing 100 3.5-oz size bars and 100 5-oz bath size bars is exist8 and exist12, respectively. (a) What is the weekly profit function P(x, y)? P(x, y) =
(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? 3.5-oz bars 5-oz bars What is the maximum weekly profit?

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Answer from: duplerk30919

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:

$\dfrac{\partial P}{\partial x} =0 \ and \ \dfrac{\partial P}{\partial y} =0$

given that:

$\dfrac{\partial^2 P}{\partial x^2} < 0 \ and \ D 0$

where;

D = Px Py -Pxy²

Now;

$\dfrac{\partial P}{\partial x} =(-0.02 x -0.01 y +72)$

$\dfrac{\partial P}{\partial y} =(-0.03 x -0.01 y +38)$

$\dfrac{\partial P}{\partial x} =0 \to 2x+y = 7200$

$\dfrac{\partial P}{\partial y} =0 \to x+3y = 3800$

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

Answer from: Quest

£56

step-by-step explanation:

Answer from: Quest

The skating rink rented 872 pairs of skates during april and may.why? : if the skating rink rented three times the pair of skates in april during may, they rented 654 skates. 218 x 3 = 654then, if they want to know how many they rented during both april and may, then you just add 654 (may) and 218 (april) to give you 872 pairs of skates for both months.654 + 218 = 872hope this !