Maximizing Profit And Investment Analysis – Economic Analysis

Price Elasticity of Demand for Good 1 and Good 2

1. Since, the interest payment is done annually, number of time interest would be paid is 5 when the bon is received at maturity (Dormady et al. 2019). Hence, the annual coupon payment of £ 80 would be received every year for a period of five years. Therefore, the annual repayment of coupon would be £ 400.
2. The annual repayment made to the investor at the end of five years is computed at £ 1400.
3. Present value of the bond due to be paid at the end of five-year lifetime is computed by adding up the present value of the face value of bond and the stream of interest payments made till maturity. Present value of interest payment is computed by adding up the present value of interest received at the end of five years as the maturity value includes both the present value of the principal amount and the interest rate. Or, the present value of bond can be obtained directly by discount the bond’s maturity value with the interest rate (Siziba and Hall 2021). It is computed that the present value of the bond to be repaid at the end of five year is £ 952.82, which implies that the bond is purchased at the higher price.
4. There exists an inverse relationship between the price of bond and market rate of interest, that is when the rate of interest declines, market price of the bond increases and vice versa (Sarwary 2020). It is clearly evident from the above calculation as when the rate of interest declines from 8% to 7.5% and 6.5%, the value of bond increases from £ 952.82 to £ 967.09 and £ 957.77 respectively. On other hand, when interest rate increased to 9%, it is observed that bond value reduced to 942.4.

 

Where P₁ and P₂ are the price of good 1 and good 2 respectively.

1. Toral revenue would represent the total revenue for the revenue earned on first and second goods respectively.
2. i) Equation for total revenue (TR)= P_1*Q₁+ P_2*Q₂=)* Q₁₊Q₂
3. ii) Profit is the difference between total revenue and total cost, hence equation for profit is given as

Profit= TR -TC

Profit=

1. TR is maximized when MR is zero

 TR₁=8Q₁ -2 Q_1* Q₁ 

TR₂= 20Q₂– 0.5*Q₂*Q₂

MR₁=8-4 Q₁

When marginal revenue is zero

Q₁=2

MR₂= 20-Q₂ 

Q₂= 20

The number of units to be sold for maximizing revenue is 2 units Q₁ of and 20 units of Q₂.

TR₁= 8*2-2*2*2=16-8=8

TR₂=20*20-0.5*20*20=400-200=200

1. The condition for maximization of profit is that its first order differentiation should be zero (Cohen et al.2019).

Hence, profit= TR-TC

Profit

Profit= 8Q₁ -2Q₁² + 20Q₂ -0.5Q₂² -10-4Q₁ -6Q₂

= 4Q₁ +14Q₂ -2Q₁² -0.5Q₂² -10

Now, ð π/ ð Q₁= 4- 4Q₁ =0

ð π/ ð Q₂ =14-Q₂= 0

Q₁ =1 and Q₂= 14

Therefore, profit is maximized when Q₁ =1 and Q₂= 14.

1. Price elasticity of demand is computed by following formula

Percentage change in quantity demanded/percentage change in price

Price when revenue is maximum that is when Q₁ =2 and Q₂= 20

P₁=8- 2Q₁= 4

P₂= 20-0.5Q₂ = 10

Price when profit is maximum that is when Q₁ =1 and Q₂= 14

P₁=8- 2Q₁= 6

P₂= 20-0.5Q₂ = 13

For good 1

Percentage change in quantity demanded= (2-1)/1.5= 67%

Percentage change in price= (2/5) *100= 40%

Price elasticity of demand for Q₁ = 1.66

For good 2

Percentage change in quantity demanded= 7/17= 41%

Percentage change in price= (3/7.7) *100= 39%

Price elasticity of demand for Q₂ = 1.05

 Answer to Question 3:

There are three variables x₁, x₂ and x₃ and the partial derivatives of f with respect to all the variables is to be found.

f (x₁) = ð f/ð x₁ = x₂ + 2 x₁ 

f (x₂) = ð f/ð x₂ = x₁ -2 x₂ x₃

f (x₃) = ð f/ð x₃ = – x²₂

1. b) Now, substituting the point (4,2,1).

f (x₁) =2+2*4=10

f (x₂) =4-2*2*1=0

f (x₃) =-(2) ^2= -4

d(f)= 10d(x₁) -4 d(x₃)

Substituting the change in value (3.8,2.1,1)

f (x₁) = 2.1 +2*3.8=9.7

f (x₂) = 3.8-2*2.1*1= -0.4

f (x₃) = -(2.1) ^2= 4.41

d(f)= 9.7d(x₁) -0.4 d(x₂) + 4.41 d(x₃)

Actual value of f = 4*2 +4^2-1*(2) ^2= 8+16-4= 20

When (4,2,1) changes to (3.8, 2.1,1)

d(x₁) =0.2

d(x₂) =-0.1

d(x₃) =0

Value of f = 10*0.2-4*0= 2

The change in the value of f when (4,2,1) changes to (3.8, 2.1,1) is 2.

1. Second order total differential

f (x₁) = ð f/ð x₁ = x₂ + 2 x₁

f (x₂) = ð f/ð x₂ = x₁ -2 x₂ x₃

f (x₃) = ð f/ð x₃ = – x²₂

Again, f is differentiated with respect to all the three variables and the second order total differentiation is given by

d(f)= x₂ + 3x₁-2 x₂ x₃– x²₂

d²(f)= 3 + (1-2 x₃-2 x₂) + (-2 x₂)

=3+1-2 x₃-2 x₂-2 x₂

=4-2 x₃-4 x₂

1. Present value of investment= £10,000

Discount rate= 15%

In order to determine the profitable investment, present value of the future cash flow of both the projects is computed using the discount rate of 15%.

Present value of CF of P1	Present value of CF of P2
1739.130435	869.5652174
1512.287335	756.1436673
1972.548697	1315.032465
1715.259737	3430.519474
1491.530206	1988.706941
8430.756409	8359.967764

Present value of net cash flows from project 1 is £ 8430.756409 and that from project 2 is £ 8359.967764. It is understood that the present value of future cash flow of project 1 is higher than project 2, which implies that project 1 is profitable than project 2. Hence, it would be advisable to invest in project 1.

1. For computing the annual interest required to grow to£ 4000 from £ 2000 in 8 years, the formula to be used is

FV=PV(1+r/n) ^n*t

Hence, r= (FV/PV) ^1/t-1

 r= (4000/2000) ^1/8-1= 2^0.125-1= 9%

1. For computing the time period required£ 2000 to grow to £ 4000, with the interest rate of 8%, the formula is

PV=FV/(1+r) ^t

2000=4000/ (1+0.08) ^t

(1+0.08) ^t = 4000/2000

(1+0.08) ^t = 2

Partial Derivatives and Optimization

1.08^t =2

Applying log on both sides

Log 2= log 1.08^t

Log 2= t*log 1.08

t= log 2/log 1.08

t= 9 years

It is observed from the above computation that when the time period of investment is less, then return is rate of return is higher and vice versa.

 Answer to Question 5:

TR= 20Q- Q²

1. Profit is expressed as the difference between the total revenue and total cost

Profit = TR -TC

Profit (π)= 20Q- Q² – (Q³– 8Q² +20Q+2)

Profit = 20Q- Q² – Q³+8Q² -20Q -2

Profit = 7Q²-Q³-2

1. Maximum profit can be found by differentiating the profit equation with respect to Q

ð π/ ð Q = 14Q-3Q²

the profit maximization point is where ð π/ ð Q= 0

14Q-3Q² = 0

14=3Q

Q= 4.67

Hence, the maximum profit would be given by 7(s4.67) ^2-(4.67) ^3-2= 48.814

1. Marginal revenue (MR)= -2Q+ 20= -2*4.67+20= 10.67

Marginal costs (MC) = 3Q² -16Q +20= 3(4.67) ^2-16(4.67) +20= 10.67

Hence, MR is equal to MC at the optimal level of output that is 4.67.

Q= 20x^0.2 y^0.8

Total expenditure = £ 1450

Price of each unit of x= £ 20

Price of each unit of y= £ 2

Expenditure incurred on strawberries= 20x

Expenditure incurred on blackberries= 2y

1. The equation of budget constraint can be given as

20x + 2y = £ 1450

1. For determining the maximum production subject to the budget constraint, the Lagrange multiplier is set up. The Lagrange multiplier for the maximum production is set up as follows

L= Q= 20x^0.2 y^0.8 + λ (725- 10x-y)

ð L/ð x = 4 x ^-0.8 y^0.8 -10 λ =0

4 x ^-0.8 y^0.8 = 10 λ ………(i)

ð L/ð y = 20 x ^0.2 y ^-0.2 -10 λ =0

16 x ^0.2 y ^-0.2 = λ …… .. (ii)

Solving (i) and (ii)

 yx^-1 = 40

y=40x

solving for x and y using the equation 10x+y=725

The value of x and y is computed at 14.5 and 580.

Hence, the maximum production is Q= 20(14.5) ^0.2*580^ (0.8) = 5546.844

1. The production is maximized at x= 14.5 and y= 580

 

It is required to show

The partial derivative of production with respect to raw material strawberries and blackberries gives the marginal product of both the raw materials (Mateer and Coppock 2018).

Q= 20x^0.2 y^0.8

MU_x = 20*0.2*x^0.2-1 * y^0.8 = 4x^-0.8 y^0.8

MU_y = 20*0.8* y^0.8-1 x^0.2 =16y^-0.2 x^0.2

In order to maximize the production, it is required to hold the following condition

MU_x/ MU_y

= y/4x

Substituting the value of x and y, it is found that

MU_x/ MU_y = 580/14.5*4= 580/58= 10

And P_x/P_y =20/2= 10

Therefore, it is observed that the production is maximized when.

 

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