# 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)
[[p=s275(3s+20), b=100000, L2=s215625(s2−1936)]]
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)
[[p=201g+925, b=100000, L3=2000001g2+4009g−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