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Now, the second derivative of the function tells us the rate of change of the first derivative. It therefore follows that if the second derivative is positive at the stationary point, then the gradient is increasing. The fact that it is a stationary point in the first place means that this can only be a minimum. Conversely, if the second derivative is negative at that point, then it is a maximum. Now, if the second derivative is zero, we have a problem. It could be a point of inflexion, or it could still be an extremum. Examples of each of these cases are below-all have a second derivative equal to zero at the stationary point in question: 1  has a point of inflexion at.
Take the derivative of the simplified equation and set it to zero in order to solve for x.. For example, if our equation was P(x) = .10x^2 - 17x - 10, the derivative set to zero would be: 0 = .20x - 17. Find the number of units you will have to produce to maximize the profit by solving for x..For example, if the derivative of our equation is 0 = .20x - 17, you would need to produce 85 units to create a maximum profit. Rearranging the equation, you get 17 = .20x, so x = 85.or example, if the derivative of our equation is 0 = .20x - 17, you would need to produce 85 units to create a maximum profit. Rearranging the equation, you get 17 = .20x, so x = 85.