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Zorluk: Çok zorLinear and Exponential Growth

An investment firm analyzes two retirement portfolios, Portfolio P1 and Portfolio P2, each established with an initial deposit of DD dollars at time t=0t = 0 years. The value of Portfolio P1 increases linearly by a constant amount each year. The value of Portfolio P2 increases exponentially at a constant annual rate. At t=10t = 10 years, the value of both portfolios is equal. At t=20t = 20 years, the value of Portfolio P2 is exactly 98\frac{9}{8} times the value of Portfolio P1. Which of the following expressions represents the value of Portfolio P2, in dollars, at t=30t = 30 years in terms of DD?

  1. A
    52D\frac{5}{2}D
  2. B
    94D\frac{9}{4}D
  3. 278D\frac{27}{8}DCevap
  4. D
    92D\frac{9}{2}D

Cevap

The expression that represents the value of Portfolio P2 at t=30t = 30 years is 278D\frac{27}{8}D.
To find the value of Portfolio P2 at t=30t = 30, we model the linear portfolio as V1(t)=D+ctV_1(t) = D + ct and the exponential portfolio as V2(t)=DgtV_2(t) = D \cdot g^t. At t=10t = 10, D+10c=Dg10D + 10c = D \cdot g^{10}, which gives c=D10(g101)c = \frac{D}{10}(g^{10}-1). At t=20t = 20, V2(20)=98V1(20)    Dg20=98(D+20c)V_2(20) = \frac{9}{8}V_1(20) \implies D \cdot g^{20} = \frac{9}{8}(D + 20c). Substituting cc into the equation and simplifying yields the quadratic equation 8(g10)218(g10)+9=08(g^{10})^2 - 18(g^{10}) + 9 = 0. Solving for g10g^{10} and selecting the value greater than 1 (since the portfolio is growing) gives g10=1.5g^{10} = 1.5. Thus, at t=30t = 30, the value of Portfolio P2 is Dg30=D(g10)3=D(1.5)3=278DD \cdot g^{30} = D \cdot (g^{10})^3 = D \cdot (1.5)^3 = \frac{27}{8}D.

Adım Adım Çözüm

1
Write the models for both portfolios in terms of the initial deposit DD.
For Portfolio P1 (linear): V1(t)=D+ctV_1(t) = D + c \cdot t, where cc is the constant annual increase.
For Portfolio P2 (exponential): V2(t)=DgtV_2(t) = D \cdot g^t, where gg is the annual growth factor.
Establishing the mathematical definitions of linear and exponential growth models is necessary to set up equations representing the given relationships.
2
Use the condition at t=10t = 10 to express the linear growth rate cc in terms of DD and g10g^{10}.
V1(10)=V2(10)    D+10c=Dg10    10c=D(g101)    c=D10(g101)V_1(10) = V_2(10) \implies D + 10c = D \cdot g^{10} \implies 10c = D(g^{10} - 1) \implies c = \frac{D}{10}(g^{10} - 1).
This allows us to eliminate the linear growth rate variable cc and express all equations in terms of the exponential growth term g10g^{10}.
3
Set up the equation for t=20t = 20 using the given relationship and substitute cc.
V2(20)=98V1(20)    Dg20=98(D+20c)V_2(20) = \frac{9}{8} V_1(20) \implies D \cdot g^{20} = \frac{9}{8} (D + 20c). Substituting cc gives:
Dg20=98(D+20D10(g101))    Dg20=98D(1+2(g101))D \cdot g^{20} = \frac{9}{8} \left( D + 20 \cdot \frac{D}{10}(g^{10} - 1) \right) \implies D \cdot g^{20} = \frac{9}{8} D \left( 1 + 2(g^{10} - 1) \right). Since D>0D > 0, we divide both sides by DD to get:
g20=98(2g101)g^{20} = \frac{9}{8} (2g^{10} - 1).
This sets up a solvable quadratic equation for the variable g10g^{10}.
4
Solve the quadratic equation for g10g^{10}.
Let x=g10x = g^{10}. The equation is x2=98(2x1)    8x218x+9=0x^2 = \frac{9}{8}(2x - 1) \implies 8x^2 - 18x + 9 = 0. Factoring the quadratic yields (2x3)(4x3)=0(2x - 3)(4x - 3) = 0, giving x=1.5x = 1.5 or x=0.75x = 0.75. Since the portfolios are growing, the growth factor g>1g > 1, so g10>1g^{10} > 1. Therefore, we choose g10=1.5g^{10} = 1.5.
Determining the value of the 10-year growth factor g10g^{10} is the key step to finding the value at future time intervals.
5
Calculate the value of Portfolio P2 at t=30t = 30 years.
V2(30)=Dg30=D(g10)3=D(1.5)3=D(32)3=278DV_2(30) = D \cdot g^{30} = D \cdot (g^{10})^3 = D \cdot (1.5)^3 = D \cdot \left(\frac{3}{2}\right)^3 = \frac{27}{8}D.
Using the properties of exponents, we express the value at 30 years in terms of the 10-year growth factor cubed.

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