In calculus, the derivative of a simple function at a given point can be thought of as the slope of a tangent line to that function at that point. We find the derivative of a function by the process of differentiation. For a simple function such as f(x) = x^a, the first derivative is given by f'(x) = a(x^(a-1)). This first derivative can be differentiated again using the exact same process to give a second derivative of the function. Both first and second derivatives are required to solve optimization problems.
For a function that produces a curved plot with finite maxima and minima, the maxima or minima of that plot will appear as peaks and valleys. At the exact top of a peak or bottom of a valley, the tangent line will be perfectly horizontal and so it will have a slope of zero. So the maxima or minima of a function will occur at those points where its derivative equals zero. To find the maxima and minima of a function, all that's needed is to take its first derivative, set the derivative to zero, then solve the resulting equation.
The first derivative only finds points that are maxima or minima, but doesn't differentiate between the two. However, as a function passes through a maximum, the rate of change in the slope of its tangent line is negative, so the second derivative of the function at a maximum will be negative. Likewise, the second derivative will be positive at a minimum.
Most optimization problems describe a physical situation for which optimization is required. Optimal conditions are found by first translating the situation into a function, then finding the value at which that function has a maximum. To find the maximum, the first derivative of the function is taken, set to zero and solved. To ensure the resulting value gives a maximum result (as opposed to a minimum), the second derivative of the function is taken and examined to ensure it is negative.
A farmer has an orchard of 50 apple trees, each of which produce 800 apples per year. If he plants more trees, he will get extra apples from those trees but each tree in the orchard above 50 lowers the apples per tree by 10. How many additional trees should be planted to get the most possible apples? If x is the number of additional trees, the total apples per year is f(x) = (50 + x)(800 - 10x) = 40,000 + 300x - 10x^2. The first derivative of this function is f'(x) = 300 - 20x, so the optimal value of x would occur when 300 - 20x = 0. Solving for x gives 15. As a double-check, the second derivative is f''(x) = -20, which is negative; thus, this is a maximum. So the farmer should plant 15 more trees to get the optimal apple harvest.