Calculus for Rational Decision Making

Prototype 1: Using differentiation

(Example of an action with negative consequences)

Consider an equation, say f(x,y) = 7x⁴ - 4y³ + x² - 8y + 5, [f(x,y) is the consequence]

x+y is the maximum "potential" of the cause [i.e. what the maximum capability to take action is, based on circumstances] - say, x+y=10.

[Here, the coefficients and powers of the variables in f(x,y) indicate how "important" they are to the result of the action.]

For "maximum benefit", either:

f(x,y)>0, f''(x,y)=0, f'''(x,y)<0 [which means that net gain is positive, and the action has the maximum rate of growth (case 1)]

or

f'(x,y)=0, f''(x,y)<0, f'(x±h,y±h)>0 (h is an independent constant) [which means that the action has the maximum net gain at present, and will continue to grow regardless of the circumstances (case 2)]

Substituting y=10-x in f(x,y) yields 7x⁴ + 4x³ - 119x² + 1208x -4075 = f(x). (The number of variables has been reduced to increase certainty.)

For no real value of x, case 1 or case 2 is satisfied. This means that the action, i.e. f(x,y), is bound to yield negative consequences. To resolve this, the coefficients and powers of variables need to be changed. (i.e. The importance given to each variable should be altered to yield maximum net gain)

Potential drawbacks:

1. Oversimplification

2. Fixed coefficients and powers

3. Non-realistic constraints

4. Complexity of solving higher-order polynomials

5. Limited scope for multivariable polynomials

6. Static analysis