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Applied calculus topic summary: applications of the derivative 
Relative Maxima and Minima
f has a relative minimum at c if there is an interval (r, s) (even a very small one) containing c for which f(c) ≤ f(x) for all x between r and s for which f(x) is defined. By a relative exremum, we mean either a relative maximum or a relative minimum. The following figure shows several relative extrema. Note Our definition of relative extremum allows f to have a relative extremum at an endpoint of its domain; the definitions used in some textbooks do not. 
Example
Here is its graph. Looking at the graph, we find that f has:


Absolute Maxima and Minima
Relative extrema may sometimes also be absolute extrema, as the following definition shows. f has an absolute maximum at c if f(c) ≥ f(x) for every x in the domain of f. f has an absolute minimum at c if f(c) ≤ f(x) for every x in the domain of f. The following figure shows several relative extrema that are also absolute extrema. Note All absolute extrema are automatically relative extrema, according to our convention. 
Example
Looking again at its relative extrema, we see that:
Note If we changed the domain to [0, +∞), then there would be no absolute maximum. (Why?) 

Locating Candidates for Relative Extrema
If f is continuous on its domain and differentiable except at a few isolated points, then its relative extrema occur among the following types of points.
The following figure shows several instances of all three types. Still uncomfortable with this material? Try the online tutorial on maxima and minima. 
Examples
1. Let us look yet again at the graph of f(x) = x^{2}  2x, with domain [0, 4].
Singular points: Let f(x) = 3(x 1)^{1/3}. Endpoints: Let f(x) = 1/x, with domain ( ∞, 0) [1, +∞).
Note If we changed the domain to [0, +∞), then there would be no absolute maximum. (Why?) 

Applications of Maxima and Minima: Optimization Problems
We solve optimization problems of the following form: Find the values of the unknowns x, y, . . . maximizing (or minimizing) the value of the objective function f, subject to certain constraints. The constraints are equations and inequalities relating or restricting the variables x, y, . . . . To solve such a problem, we use the constraint equations to write all of the variables in terms of one chosen variable, substitute these into the objective function f, and then find extrema as above. (We use any constraint inequalities to determine the domain of the resulting function of one variable.) Specifically:
1. Identify the unknown(s).
2. Identify the objective function.
3. Identify the constraint(s).
4. State the optimization problem.
5. Eliminate extra variables.
6. Find the absolute maximum (or minimum) of the objective function.

Example
Here is a maximization problem:
Let us carry out the procedure for solving. Since we already have the problem stated as an optimization problem, we can start at Step 5.
(100 2y) ≥ 0, or y ≤ 50. 6. Find the absolute maximum (or minimum) of the objective function.
From the table, we see that the largest possible value of A is 1,250, which occurs when y = 25. The corresponding value of x is x = 100  2y, so x = 50 when y = 25. 

Acceleration, Concavity, and the Second Derivative
Acceleration
Concavity

Examples
Acceleration
Acceleration = a(t) = s" (t) = 6t + 4 miles per hour per hour. Concavity
f"(x) = 6x is negative when x < 0 and positive when x > 0. The graph of f is concave down when x < 0 and concave up when x > 0. f has a point of inflection at x = 0, where the second derivative is 0. 

Analyzing Graphs
We can use graphing technology to draw a graph, but we need to use calculus to understand what we are seeing. The most interesting features of a graph are the following. Features of a Graph
2. Relative extrema Use the techniques described above to locate the relative extrema. 3. Points of inflection Set f"(x) = 0 and solve for x to find candidate points of inflection. 4. Behavior near points where the function is not defined If f(x) is not defined at x = , consider lim_{x → } f(x) and lim_{x → +} f(x) to see how the graph of f approaches this point. 5. Behavior at infinity Consider lim_{x → ∞} f(x) and lim_{x → +∞} f(x) if appropriate, to see how the graph of f behaves far to the left and right. If you have Excel, try our Excel First and Second Derivative Graphing Utility to see plots of any function and its first two derivatives. 
Example
Here is the graph of
To analyze this, we follow the procedure at left: 1. The x and yintercepts Setting y = 0 and solving for x gives x = 0. This is the only xintercept. Setting x = 0 and solving for y gives y = 0: the yintercept. 2. Relative extrema The only extrema are stationary points found by setting f'(x) = 0 and solving for x, giving x = 0 and x =  4. The corresponding points on the graph are the relative maximum (0, 0) and the relative minimum at ( 4, 8/9). 3. Points of inflection Solving f"(x) = 0 analytically is difficult, so we can solve it numerically (plot the second derivative and estimate where is crosses the xaxis) and find that the point of inflection lies at x  6.1072. 4. Behavior near points where the function is not defined The function is not defined at x =  1 and x = 2. The limits as x approaches these values from the left and right can be inferred from the graph:
5. Behavior at infinity From the graph (or the function) we see that


Related Rates
If Q is a quantity that is varying with time, then the rate at which Q is changing is given by the time derivative, dQ/dt. A typical related rates problem asks for the rate of change of a quantity Q, given the rates of changes of various other related quantities. Solving a Related Rates Problem A. The Problem
B. The Relationship
C. The Solution
Take a look at the online tutorial on related rates. 
Example
The traffic at the RealWorld web site is given by


Elasticity of Demand
The elasticity of demand, E, is the percentage rate of decrease of demand per percentage increase in price. We obtain it from the demand equation according to the following formula:
where the demand equation expresses demand, q, as a function of unit price, p. We say that the demand is elastic if E > 1, the demand is inelastic if E < 1, and the demand has unit elasticity if E = 1. To find the unit price that maximizes revenue, we express E as a function of p, set E = 1, and then solve for p. 
Example
Suppose that the demand equation is q = 20,000  2p. Then
If p = 2,000, then E = 1/4, and demand is inelastic at this price. If p = 8,000, then E = 4, and demand is elastic at this price. If p = 5,000, then E = 1, and the demand has unit elasticity at this price. 