An analyst...(continue reading)?

Question

An analyst has pedicted that the growth rate, as a percent, for a specific mutual fund can be modelled by the equation:

r = 0.5x^3/2 - 6x^2/3 + 10.1, where x is the number of months and x is an element of [0,24]. Determine the best and worst times to invest in this mutual fund.

Answer 1

To determine the best and worst times to invest, we need to find maximum value and minimum value.
We have to find dr/dx = 0 where the gradient show maximum or minum value.

r = 0.5x^3/2 - 6x^2/3 +10.1
dr/dx = 0.5(3/2)x^1/2 - 6(2/3)x^(-1/3)
dr/dx = 0
0.5(3/2)x^1/2 - 6(2/3)x^(-1/3) = 0
3/4x^1/2 - 4x^(-1/3) = 0
3/4x^1/2 = 4x^(-1/3)
x^5/6 = 16/3 = 5.33
x = 5.33^6/5
x = 7.45

To find maximum or minimum of r value, we have to find d^2r/dx^2
dr/dx = 3/4x^1/2 - 4x^(-1/3)
d^r/dx^2 = 3/4(1/2)x^(-1/2) -4(-1/3)x^(-4/3)
d^r/dx^2 = 3/8x^(-1/2)+4/3x^(-4/3)

For x = 7.45
d^r/dx^2 = 3/8(7.45)^(-1/2)+4/3(7.45)^(-4/3) = +positive
So, r is minimum

As x between 0, 24, the maximum value of r is either at x=0 or x =24
if x = 0, r=0.5(0)^3/2 - 6(0)^2/3 +10.1 = 10.1(starting)

if x = 24,
r = 0.5(24)^3/2 - 6(24)^2/3 +10.1 = 18.97 (maximum)

if x =7.45 ,
r = 0.5(7.45)^3/2 - 6(7.45)^2/3 +10.1 = -2.62 (minimum)

So the best time to invest is when the graph in lowest point (x =7.45) and the rate start picking up
and the worst time to invest at x=0 because the graph is dropping...

Answer 2

you have to find the minimum and maximum values of r, as x varies betweeen 0 and 24

1) differentiate with reaspect to x
2) solve r' = 0 , deterimine if solutions these are min and max

Answer 3

diffrentiate the eqn wrt x

you will get the slope.

put the values and get the value of slope.

maxima and minima can be found by diffretiating again.