(a) Solve the problem to maximise the utility, the constraint by the budget using Lagrange multipliers. Step 1: Initialize the Lagrange function: L(x, y, ) = (13x + 15y - 560) x(y + 1) Step 2: Take the partial derivatives and multiply them by zero: ¿L/Öx = y + 1 - 13A = 0 (1) ¿L/By = x - 15A = 0 (2) 13x + 15y - 560 = 0 (3) Step 3: Concurrently solve the systems of equations (1), (2), and (3): x = 15à in equation (2). When we substitute x in equation (3), we get: 13(15) + 15y - 560=0 195A+15y - 5600 15y=560-195 (4) Substituting x = 15 and equation (4) into equation (1) yields: y +1-13=0y=13-1 (5) Substituting equation (5) for equation (4): 15(13-1)=560-195 195A-15=560-195 3901=575-575/390 A = 1.474358974 1.474 Substituting 1.474 into equation (5) yields: y=13(1.474)-1 y=18.16666667 18.167 Substituting x = 15A into equation (2) yields: x = 15(1.474) x=22.11538462 = 22.11 As a result, the best values are x = 22.11 & y = 18.167. (b) Show that the solution found does actually maximise the utility function (amongst feasible x and y choices). To demonstrate that the solution identified maximises the utility function, we must first determine whether it meets the second- order requirements for a maximum. This necessitates computing and evaluating the second-order partial derivatives of the Lagrang function at optimum values. #Liêx* =Ũ ởLiêyt = 0 ởL/©xây=1 We may deduce that the solution maximises the utility function since the second-order partial derivatives are constant and do not rely on x or y. (e) Calculate the optimal values for x, y, à, and U. Determine the best values for x, y, and U: We get x = 22.11538462 22.11 y = 18.16666667 18.167. λ=1474358974 1.474 U( x, y) = x (y+1 ) = 22.11( 18.167 +1) = 423.88 (d) Consider now the dual optimisation problem of minimising the cost (Budget, B(x, y)) to achieve the utility U as calculated in (c). Relate the optimal values of this problem to those found in (e) and also B.. The goal of the dual optimisation problem is to minimise the cost (budget) B(x, y) in order to maximise the utility U as estimated in (c). Because we now know the best values for x and y, we can plug them into the budget constraint equation:B 13(22.11538462) + 15(18.16666667) B 287.43+272.43-B B=560 The best solution for the dual optimisation issue is B=560, which equals the entire budget.
(a) Solve the problem to maximise the utility, the constraint by the budget using Lagrange multipliers. Step 1: Initialize the Lagrange function: L(x, y, ) = (13x + 15y - 560) x(y + 1) Step 2: Take the partial derivatives and multiply them by zero: ¿L/Öx = y + 1 - 13A = 0 (1) ¿L/By = x - 15A = 0 (2) 13x + 15y - 560 = 0 (3) Step 3: Concurrently solve the systems of equations (1), (2), and (3): x = 15à in equation (2). When we substitute x in equation (3), we get: 13(15) + 15y - 560=0 195A+15y - 5600 15y=560-195 (4) Substituting x = 15 and equation (4) into equation (1) yields: y +1-13=0y=13-1 (5) Substituting equation (5) for equation (4): 15(13-1)=560-195 195A-15=560-195 3901=575-575/390 A = 1.474358974 1.474 Substituting 1.474 into equation (5) yields: y=13(1.474)-1 y=18.16666667 18.167 Substituting x = 15A into equation (2) yields: x = 15(1.474) x=22.11538462 = 22.11 As a result, the best values are x = 22.11 & y = 18.167. (b) Show that the solution found does actually maximise the utility function (amongst feasible x and y choices). To demonstrate that the solution identified maximises the utility function, we must first determine whether it meets the second- order requirements for a maximum. This necessitates computing and evaluating the second-order partial derivatives of the Lagrang function at optimum values. #Liêx* =Ũ ởLiêyt = 0 ởL/©xây=1 We may deduce that the solution maximises the utility function since the second-order partial derivatives are constant and do not rely on x or y. (e) Calculate the optimal values for x, y, à, and U. Determine the best values for x, y, and U: We get x = 22.11538462 22.11 y = 18.16666667 18.167. λ=1474358974 1.474 U( x, y) = x (y+1 ) = 22.11( 18.167 +1) = 423.88 (d) Consider now the dual optimisation problem of minimising the cost (Budget, B(x, y)) to achieve the utility U as calculated in (c). Relate the optimal values of this problem to those found in (e) and also B.. The goal of the dual optimisation problem is to minimise the cost (budget) B(x, y) in order to maximise the utility U as estimated in (c). Because we now know the best values for x and y, we can plug them into the budget constraint equation:B 13(22.11538462) + 15(18.16666667) B 287.43+272.43-B B=560 The best solution for the dual optimisation issue is B=560, which equals the entire budget.
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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
Transcribed Image Text:(a) Solve the problem to maximise the utility, the constraint by the budget using Lagrange multipliers.
Step 1: Initialize the Lagrange function:
|L(x, y, ) = (13x + 15y - 560) x(y + 1)
Substituting x= 15 and equation (4) into equation (1)
yields: y + 1.13=0y=13-1 (5)
Step 2: Take the partial derivatives and multiply them by zero:
Substituting equation (5) for equation (4):
15(13A-1)=560-195
OL/0x=y+ 1-13 -0 (1)
19515 560-195
3901 = 575=575/390
| ÔL/Öy = x - 15A = 0 (2)
λ = 1.474358974 1.474
13x + 15y - 560 = 0 (3)
Substituting 1.474 into equation (5) yields:
y = 13(1.474) -1
Step 3: Concurrently solve the systems of equations (1), (2), and (3): y = 18.16666667 18.167
|x = 15A in equation (2).
Substituting x = 15A into equation (2) yields:
When we substitute x in equation (3), we get:
13(15) + 15y - 560 = 0 195A + 15y - 560 = 0 15y = 560 - 195A (4)
x = 15(1.474)
x = 22.11538462 ≈ 22.11
As a result, the best values are x = 22.11 & y = 18.167.
(b) Show that the solution found does actually maximise the utility function (amongst feasible x and y choices).
To demonstrate that the solution identified maximises the utility function, we must first determine whether it meets the second-
order requirements for a maximum. This necessitates computing and evaluating the second-order partial derivatives of the Lagrange
function at optimum values.
PLiêx* = 0 #L/Ðy! = 0 #L/êxây=1
We may deduce that the solution maximises the utility function since the second-order partial derivatives are constant and do not
rely on x or y.
(e) Calculate the optimal values for x, yº, λº, and U-
Determine the best values for x, y, and U: We get x = 22.11538462 = 22.11 y = 18.16666667 18.167. A = 1474358974 1.474
|U( x, y) = x (y+1 ) = 22.11(18.167 +1) = 423.88
(d) Consider now the dual optimisation problem of minimising the cost (Budget, B(x, y)) to achieve the utility U- as
calculated in (c). Relate the optimal values of this problem to those found in (c) and also B..
The goal of the dual optimisation problem is to minimise the cost (budget) B(x, y) in order to maximise the utility U as estimated in
Because we now know the best values for x and y, we can plug them into the budget constraint equation:B 13(22.11538462) +
15(18.16666667)=B 287.43 +272.43 =B
B 560
The best solution for the dual optimisation issue is B = 560, which equals the entire budget.

Transcribed Image Text:Question 2
Steven has struck a deal with his dad to buy his car when he can afford to.
The car is valued at $55 000 today but depreciates at a continuously compounding rate of
1% per month (i.e. 1 — d = e−0.01).
Steven has $9 000 in a bank account and plans to add $120 each month end. The bank
pays interest at a continuously compounding rate of 1% per month (i.e. 1 + r = 0.0¹).
(a) Formulate the value of the car as a finite difference equation and solve by calculating
the Complementary Function and Particular Solution.
(b) Formulate Steven's Savings amount in a similar way and solve.
(c) Solve to equate the values in (a) and (b) to find the time when Steven can buy the
car.
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