Answer:
a-1. We have:
MC = 3y^2 / 800 <=== Marginal cost (MC) function
AC = (200 / y) + (y^2 / 800) <=== Average cost (AC) function
a-2. The amount of output is 43.09 bouquets.
a-3. Average cost at this level is $6.96 per unit.
b-1. We have:
MC = 3y^2 / 2,000 <=== Marginal cost (MC) function
AC = (500 / y) + (y^2 / 2,000) <=== Average cost (AC) function
b-2. The amount of output is 79.37 bouquets.
b-3. Average cost at this level is $9.45 per unit.
c-1. We have:
MC = 3y^2 / 4,000 <=== Marginal cost (MC) function
AC = (1,000 / y) + (y^2 / 4,000) <=== Average cost (AC) function
c-2. The amount of output is 125.99 bouquets.
c-3. Average cost at this level is $11.91 per unit.
Explanation:
Given:
cv(y) = y^3/ 4F ………………… (1)
cf = fixed cost = F
Therefore, total cost (C) per month is as follows:
C(y) = cf + cv(y) = y^ 3/ 4F
C(y) = F + y^3 / 4F ……………………… (2)
a-1. If she has 200 square feet of floor space, write down her marginal cost function and her average cost function.
This implies that:
F = 200
Marginal cost (MC) function is obtained by taking the first derivative of equation (1) and substituting F = 200 as follows:
MC = cv’(y)
MC = 3y^2 / (4 * 200)
MC = 3y^2 / 800 ………………. (3) <= Marginal cost (MC) function
Average cost (AC) function can be obtained by dividing equation (2) by y, substituting F = 200 and solve as follows:
AC = C’(y) = (200 / y) + (y^3 / 4F) / y
AC = (200 / y) + (y^3 / (4 * 200)) / y
AC = (200 / y) + (y^2 / 800) …………………. (4) <= Average cost (AC) function
a-2. At what amount of output is average cost minimized?
Since average cost is minimized when MC = AC, we therefore equate equations (3) and (4) and solve for y as follows:
3y^2 / 800 = (200 / y) + (y^2 / 800)
0.00375y^2 = (200 / y) + 0.00125y^2
0.00375y^2 - 0.00125y^2 = 200 / y
0.0025y^2 = 200 / y
(0.0025y^2)y = 200
0.0025y^3 = 200
y^3 = 200 / 0.0025
y^3 = 80,000
y = 80,000^(1/3)
y = 43.09
Therefore, the amount of output at which is average cost minimized is 43.09 bouquets.
a-3. At this level of output, how much is average cost?
Substituting y = 43.09 into equation (4), we have:
AC = (200 / 43.09) + (43.09^2 / 800)
AC = 6.96
Therefore, average cost at this level is $6.96 per unit.
b-1. If she has 500 square feet, write down her marginal cost function and her average cost function.
This implies that:
F = 500
Marginal cost (MC) function is obtained by taking the first derivative of equation (1) and substituting F = 500 as follows:
MC = cv’(y)
MC = 3y^2 / (4 * 500)
MC = 3y^2 / 2,000 ………………. (5) <= Marginal cost (MC) function
Average cost (AC) function can be obtained by dividing equation (2) by y, substituting F = 500 and solve as follows:
AC = C’(y) = (500 / y) + (y^3 / (4 * 500)) / y
AC = (500 / y) + (y^3 / (4 * 500)) / y
AC = (500 / y) + (y^2 / 2,000) …………………. (6) <= Average cost (AC) function
b-2. At what amount of output is average cost minimized?
Since average cost is minimized when MC = AC, we therefore equate equations (5) and (6) and solve for y as follows:
3y^2 / 2,000 = (500 / y) + (y^2 / 2,000)
0.0015y^2 = (500 / y) + 0.0005y^2
0.0015y^2 - 0.0005y^2 = 500 / y
0.001y^2 = 500y
0.001y^2 * y = 500
0.001y^3 = 500
y^3 = 500 / 0.001
y^3 = 500,000
y = 500,000^(1/3)
y = 79.37
Therefore, the amount of output at which is average cost minimized is 79.37 bouquets.
b-3. At this level of output, how much is average cost?
Substituting y = 79.37 into equation (6), we have:
AC = (500 / 79.37) + (79.37^2 / 2,000)
AC = 9.45
Therefore, average cost at this level is $9.45 per unit.
c-1. If she has 1,000 square feet, write down her marginal cost function and her average cost function.
This implies that:
F = 1,000
Marginal cost (MC) function is obtained by taking the first derivative of equation (1) and substituting F = 1,000 as follows:
MC = cv’(y)
MC = 3y^2 / (4 * 1,000)
MC = 3y^2 / 4,000 ………………. (7) <= Marginal cost (MC) function
Average cost (AC) function can be obtained by dividing equation (2) by y, substituting F = 1,000 and solve as follows:
AC = C’(y) = (1,000 / y) + (y^3 / (4 * 1,000)) / y
AC = (1,000 / y) + (y^3 / (4,000)) / y
AC = (1,000 / y) + (y^2 / 4,000) …………………. (8) <= Average cost (AC) function
c-2. At what amount of output is average cost minimized?
Since average cost is minimized when MC = AC, we therefore equate equations (7) and (8) and solve for y as follows:
3y^2 / 4,000 = (1,000 / y) + (y^2 / 4,000)
0.00075y^2 = (1,000 / y) + 0.00025y^2
0.00075y^2 - 0.00025y^2 = 1,000 / y
0.0005y^2 = 1,000 / y
0.0005y^2 * y = 1,000
y^3 = 1,000 / 0.0005
y^3 = 2,000,000
y = 2,000,000^(1/3)
y = 125.99
Therefore, the amount of output at which is average cost minimized is 125.99 bouquets.
c-3. At this level of output, how much is average cost?
Substituting y = 125.99 into equation (8), we have:
AC = (1,000 / 125.99) + (125.99^2 / 4,000)
AC = 11.91
Therefore, average cost at this level is $11.91 per unit.
