2017-10-14-215929.sagews

# Variables

# p = price of the manufacture
# b = budget for advertising
# To = total of unit sold
# R = Revenue
# C= cost
# P = Profit

# Assumptions
# R(p,b) = To(p,b) * p
# To(p,b) = 10000 + 200*(b-50000)/10000 + 5000*(950-p)/100
# C(p,b) = 700 * To(p,b) + b
# P(p,b) = R(p,b) - C(p,b)

# Constraints
# 0 <= p
# 50000 <= b
# b <= 100000
To(p,b) = 10000 + 200*(b-50000)/10000 + 5000*(950-p)/100
R(p,b) = To(p,b) * p
C(p,b) = 700 * To(p,b) + b
P(p,b) = R(p,b) - C(p,b)

p1 = contour_plot(P(p,b),(p,-10000,10000),(b,50000,100000))
p2 = implicit_plot(p==0, (p,-10000,10000),(b,50000,100000))
plot(p1+p2)

# When gradient = 0

solve([diff(P,p)==0, diff(P,b)==0],p,b)

[[p == 750, b == -825000]]

# constraint p = 700
L1 = var('L1')
# constraint p = 700
solve([diff(P,p)==L1,diff(P,b)==0,p==700],p,b,L1)

[]

# constraint p = 950
solve([diff(P,p)==L1,diff(P,b)==0,p==950],p,b,L1)

[]

# constraint b = 50000
solve([diff(P,p)==0,diff(P,b)==L1,b==50000],p,b,L1)

[[p == 925, b == 50000, L1 == (7/2)]]

# constraint b = 100000
solve([diff(P,p)==0,diff(P,b)==L1,b==100000],p,b,L1)

[[p == 935, b == 100000, L1 == (37/10)]]

# Test of Points
P(0,50000)
P(0,100000)
P(925,50000)
P(935,100000)

-40300000 -41050000 2481250 2661250

# The manufacturer of the personal computer reachs the maximum profit when he balances advertising to $100000 and price to $935

#b Determine the sensitivity of the decision variables to the price

To(p,b,s) = 10000 + 200*(b-50000)/10000 + s*(950-p)
R(p,b,s) = To(p,b,s) * p
C(p,b,s) = 700 * To(p,b,s) + b
P(p,b,s) = R(p,b,s) - C(p,b,s)

# constraint b = 100000
L2 = var('L2')
eq = solve([diff(P,p)==0,diff(P,s)==L2, b==100000],p,b,L2)

show(eq)

[[$\displaystyle p = \frac{275 \, {\left(3 \, s + 20\right)}}{s}$, $\displaystyle b = 100000$, $\displaystyle L_{2} = \frac{15625 \, {\left(s^{2} - 1936\right)}}{s^{2}}$]]

Sp(s) = diff(eq[0][0].rhs())*s/(eq[0][0].rhs())

n(Sp(50))

-0.117647058823529

# Compute the sensitivity of the decision variable to the advertising agency's estimateof 200 new sales each time

To(p,b,g) = 10000 + g*(b-50000)/10000 + 5000*(950-p)/100
R(p,b,g) = To(p,b,g) * p
C(p,b,g) = 700 * To(p,b,g) + b
P(p,b,g) = R(p,b,g) - C(p,b,g)

L3 = var('L3')
eq = solve([diff(P,p)==0,diff(P,b)==L3,b==100000],p,b,L3)

show(eq)

[[$\displaystyle p = \frac{1}{20} \, g + 925$, $\displaystyle b = 100000$, $\displaystyle L_{3} = \frac{1}{200000} \, g^{2} + \frac{9}{400} \, g - 1$]]

Sg(g) = diff(eq[0][0].rhs())*g/(eq[0][0].rhs())

n(Sg(200))

0.0106951871657754

# The value of the multiplier is L1 in the question (a) is 3.7. it is representative of the monetary value of advertising. It means that if the manufacturer increases $1 in his budget for advertising, he will increase his profit by $3.70