An individual has $40,000 to invest: $28,000 will be put into a low-risk mutual fund averaging 6.9% interest compounded monthly, and the rem

Question

An individual has $40,000 to invest: $28,000 will be put into a low-risk mutual fund averaging 6.9% interest compounded monthly, and the remainder will be invested in a high-yield bound fund averaging 9.8% interest compounded continuously.

Required:
a. Write an equation for the total amount in the two investments.
b. Write the rate-of-change equation for the combined amount.
c. How rapidly is the combined amount of the investments growing after 6 months? after 15 months?

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MichaelMet 6 months 2021-08-22T15:07:50+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-08-22T15:09:16+00:00

    Answer:

    a) F(x) = 28,000( 1.00575 )^12x  + 12,000e^0.098x

    b)  F'(x ) = 28,000 ( In 1.071224 ) ( 1.071224 )^x + 1176 e^0.098x  dollar per year

    c) 3228.94 dollar/year,   3428.73 dollar/year

    Step-by-step explanation:

    Capital = $40,000

    $28,000 = low-risk mutual fund

    6.9% monthly compounded interest for the low risk mutual fund

    $12,000  = high-risk yield bound fund

    9.8% continuously compounded interest

    A) Equation for total amount in two investments

    F(x) = F1(x) + F2(x) —– ( 1 )

    where :

    F1(x) ( future value for monthly compounded interest)

    =  28,000( 1 + 0.069/12 )^12x = 28,000 ( 1.00575 )^12x

    F2(x) ( future value for continuously compounded interest )

    = ( 40,000 – 28,000 )e^0.098x = 12,000 e^0.098x

    back to equation 1

    F(x) = 28,000( 1.00575 )^12x  + 12,000e^0.098x

    B Rate of change equation

    f'(x) = d/dx (28,000( 1.00575 )^12x) + d/dx ( 12,000e^0.098x )

    ∴ f'(x) = 28,000 d/dx (1.00575^12)^x + 12,000 d/dx(b^x)

             = 28,000 ( In 1.071224 ) ( 1.071224 )^x + 12,000 ( In b ) ( b^x )

      f'(x ) = 28,000 ( In 1.071224 ) ( 1.071224 )^x + 1176 e^0.098x  dollar per year

    C) Determine how rapidly the combined amount will grow after 6 months and after 15 months

    i.e. x = 0.5 , x = 1.25 years

    after 6 months

    28,000 ( In 1.071224 ) ( 1.071224 )^(0.5) + 1176*e^((0.098(0.50))

    =  1993.88  + 1235.06 = 3228.94 dollar/year

    after 15 months

    28,000 ( In 1.071224 ) ( 1.071224 )^(1.25) + 1176*e^((0.098(1.25 ))

    =   2099.47  +  1329.26 = 3428.73 dollar/year

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