Economics Assignment: Linear Functions, Derivatives, and Present Value

This assignment provides solutions to several economics problems. The first problem involves determining the equation of a linear function representing cost based on the number of copies made. The second problem uses derivatives to find the maximum number of households using a product. The third problem calculates the time required for cell growth to reach a certain level using a linear function. The fourth problem determines the number of months required for an investment to reach a specific value using compound interest. Finally, the fifth problem calculates the present value of an annuity and then determines the annual payment required to achieve the same future value with a different interest rate. Desklib provides a platform for students to access these solved assignments and other study resources.

1)
This is a linear function of the type y=mx + b, where the y-intercept (b) is the setup cost
and the slope (m) is the cost per copy
you have two points on the line, (100, 89) and (200, 93)
so you can find the equation
First, the slope, m = (93-89)/ (200-100) => m = 4/100 => m = .04
so each copy costs 4 cents
substituting 89 = .04(100) + b => 89 = 4 + b => b = 85
Cost = ( 0.04 ) × pages +85
2)
f ( x )=x ( 12−x ) where 0< x< 12
Now to find the maximum we need to find the derivative of f ( x )
d ( f ( x ) )
dx =12−2 x=0=¿ x =6
So maximum number of household using this product will be obtained by putting x=6
So,
f ( x ) =6 ( 12−6 ) =36
Therefore 36 thousands household will be using this product.
3)
This is a linear function of the type y=mx + b. From the question m = 0.07, Need to find out t such
that cells ≥ 5000
C ells= ( 0.07 ) × time+2000
( 0.07 ) × time+2000 ≥ 5000
( 0.07 ) × time ≥ 3 000
time ≥ 3 000
0.07
time=42858 hrs
4)

So, monthly interest is = 6/12 = 0.5%
10000=1000 (1+0.005)n
10000=1000 (1.005)n
10=(1.005)n
Taking log on both sides we get,
n=461.66 months=38.47 years
5)
Present Value is given by,
PV =P [1− ( 1+r )−n ]
n
Putting P = $60,000, n = 8 years and r = 6%, we get PV as,
PV = $372,587.63 rounded to 2 decimal places.
Now when I want the return to be 4 %, then the amount to be given per year for the FV =
$372,587.63 purchase price:
FV = 372587.63
n = 8
payments are set to be made at the end of each time period P,
F V =P [ ( 1+r ) n −1]
n
Putting the values we get, P = $40.436.13