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The function
[tex]\[ A(x) = P(1 + b x)^m (1 - x)^n \][/tex]
models the final value [tex]\( A(x) \)[/tex] of a bankroll after [tex]\( N = m + n \)[/tex] consecutive wagers on an event with given odds [tex]\( b \)[/tex] to 1 against.

In this model, the true odds are [tex]\( m \)[/tex] to [tex]\( n \)[/tex] in favor of the event. In other words, [tex]\( m \)[/tex] represents the number of wins, [tex]\( n \)[/tex] the number of losses, and [tex]\( N = m + n \)[/tex] is the total number of wagers. The value [tex]\( P \)[/tex] represents the initial bankroll and [tex]\( x \)[/tex] represents the fraction of the bankroll wagered at each of the [tex]\( N \)[/tex] consecutive events.

Suppose the initial bankroll is [tex]\( \$ 4100 \)[/tex], the true number of wins is [tex]\( m = 33 \)[/tex], the true number of losses is [tex]\( n = 14 \)[/tex], and [tex]\( b = 1.44 \)[/tex].

Find the value of [tex]\( x \)[/tex] that will maximize [tex]\( A(x) \)[/tex] on the interval [tex]\([0,1]\)[/tex].

The maximum value of [tex]\( A(x) \)[/tex] is [tex]\(\square\)[/tex],
which occurs at [tex]\( x = \square \)[/tex].



Answer :

GinnyAnswer
To determine the value of [tex]\( x \)[/tex] that maximizes the function [tex]\( A(x) = P(1 + b x)^m (1 - x)^n \)[/tex], where:
- [tex]\( P = 4100 \)[/tex] is the initial bankroll,
- [tex]\( b = 1.44 \)[/tex],
- [tex]\( m = 33 \)[/tex] is the number of wins,
- [tex]\( n = 14 \)[/tex] is the number of losses,
- [tex]\( x \)[/tex] is the fraction of the bankroll wagered at each of the [tex]\( N = m + n = 47 \)[/tex] consecutive events,

we proceed as follows:

1. Rewrite the function with the given values:
[tex]\[ A(x) = 4100 (1 + 1.44 x)^{33} (1 - x)^{14} \][/tex]

2. Identify the optimization problem:
We need to find the value of [tex]\( x \)[/tex] on the interval [tex]\([0, 1]\)[/tex] that maximizes the function [tex]\( A(x) \)[/tex].

3. Optimize the function:
The optimal value of [tex]\( x \)[/tex] that maximizes [tex]\( A(x) \)[/tex] can be found using numerical methods to determine the peak value within the domain [tex]\( 0 \leq x \leq 1 \)[/tex].

The maximum value of [tex]\( A(x) \)[/tex] occurs at:
[tex]\[ x \approx 0.49527189227236057 \][/tex]

4. Calculate the maximum value of [tex]\( A(x) \)[/tex]:
Substitute [tex]\( x = 0.49527189227236057 \)[/tex] back into the function to find [tex]\( A(x) \)[/tex]:
[tex]\[ A \left( 0.49527189227236057 \right) \approx 14831463.175097775 \][/tex]

Therefore, the maximum value of [tex]\( A(x) \)[/tex] is approximately [tex]\( 14831463.175097775 \)[/tex], which occurs at [tex]\( x \approx 0.49527189227236057 \)[/tex].

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