1. Managerial Economics
Managerial Economics = economic theory + mathematical eco statistical analysis

2. FUNCTIONAL RELATIONSHIPS AND ECONOMIC MODELS
In mathematics,
y = f (x)
is read “y is a function of x.
This relationship indicates that the value of y depends in a systematic way on the value of x.
The variable y is referred to as the dependent variable. The variable x is referred to as the independent variable.

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Diagramatic Representation

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The functional notation f in above equation can be regarded as a specific rule that defines the relationship between values of x and y.
For example, that y = f(x) = 3x, the actual rule has been made explicit.
In this case, the rule asserts that when x = 2, then y = 6, and so on.
In this case, the value of x has been transformed, or mapped, into a value of y.

5. METHODS OF EXPRESSING ECONOMIC AND BUSINESS RELATIONSHIPS
Economic and business relationships may be represented in a variety of ways, including tables, charts, graphs, and algebraic expressions.
For example the following equation summarizes total revenue (TR) for a
firm operating in a perfectly competitive industry.
TR = f (Q) = PQ

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Suppose that the selling price is $18.The total revenue function of the firm may be expressed algebraically as
TR = 18Q
For the output levels Q = 0 to Q =6, this relationship may be expressed in tabular form

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Total revenue for the output levels Q = 0 to Q = 6 may also be expressed diagrammatically.

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The total revenue (TR) in Figure illustrates the general class of mathematical relationships called linear functions. A linear function may be written in the general form
y = f (x) = a + bx
where a and b are known constants. The value a is called the intercept and the value b is called the slope.

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Mathematically, the intercept is the value of y when x = 0. This expression is said to be linear in x and y, where the corresponding graph is represented by a straight line.
In the total revenue example just given, a = 0 and b = 18.

11. THE SLOPE OF A LINEAR FUNCTION
In Equation, the parameter b is called the slope, which obtained by dividing the change in the value of the dependent variable (the “rise”) by the change in the value of the independent variable (the “run”) as we move between two coordinate points. The value of the slope may be calculated by using the equation

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In terms of our total revenue example, consider the quantity total revenue combinations (Q1, TR1) = (1, 18) and (Q2, TR2) = (3, 54). In this case, a measure of the slope is given by the expression

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which, in this case, is the price of the product.

14. ECONOMIC OPTIMIZATION
Many problems in economics involve the determination of “optimal” solutions. For example, a decision maker might wish to determine the level of output that would result in maximum profit.
The process of economic optimization essentially involve three steps:
1. Defining the goals and objectives of the firm
2. Identifying the firm’s constraints

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3. Analyzing and evaluating all possible alternatives available to the decision maker
In essence, economic optimization involves maximizing or minimizing some objective function, which may or may not be subject to one or more constraints.

16. OPTIMIZATION ANALYSIS
The process of economic optimization may be illustrated by considering the firm’s profit function p, which is defined as
TR = 18Q
TC = Q- 9Q2 +Q3

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The extreme values can be read directly from Table and Figure, they also may be determined directly from the underlying p function.

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For a clue to how this might be determined, note that the slope (steepness) of the p curve at both the minimum and maximum output levels would be zero; that is, the curve would be neither upward sloping nor downward sloping.
In the case of the total revenue equation, for example, the slope at any output level is defined as the “rise” over the “run,” that is, change in total revenue divided by the change in output, DTR/DQ. But this is the definition of marginal revenue (MR).

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Fortunately, differential calculus offers an easy way to find the marginal function by taking the first derivative of the total function.
In the case of our example, total profit will be maximized or minimized at an output level at which the slope of the profit function is zero.
This is accomplished by finding the first derivative of the profit function, setting it equal to zero, and solving for the value of the corresponding output level.

21. DERIVATIVE OF A FUNCTION
Consider, again, the function 
y = f (x)
The first derivative of a function (dy/dx) is simply the slope of the function when the interval along the horizontal axis (between x1 and x2) is made infinitesimally small.
Technically, the derivative is the limit of the ratio Dy/Dx as Dx approaches zero, that is,

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Calculus offers a set of rules for using derivatives (slopes) for making optimizing decisions such as minimizing cost (TC) or maximizing total profit (p).

23. RULES OF DIFFERENTIATION
POWER-FUNCTION RULE
A power function is of the form
where a and b are real numbers. The rule for finding the derivative of a power function is
Example

24. SUMS AND DIFFERENCES RULE
If we define g and h to be functions of the variable x, then

25. PRODUCT RULE Further, let
if we define g and h to be functions of the variable x, then
Further, let
The derivative of a product is defined as

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28. QUOTIENT RULE
Again defining g and h as functions of x, we write

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30. CHAIN RULE
Let y = f(u) and u = g(x). Substituting, we are able to write the composite function

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32. TOTAL, AVERAGE, AND MARGINAL RELATIONSHIPS

33. PROFIT MAXIMIZATION: THE FIRST-ORDER CONDITION
We are now in a position to use the rules for taking first derivatives to find the level of output Q that maximizes p. Consider again the total revenue and total cost functions introduced earlier

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The first-order condition for profit maximization is dp/dQ = 0, thus
In this example, therefore, the entrepreneur of the firm would maximize his profits at Q = 5.

35. PROFIT MAXIMIZATION: THE SECOND-ORDER CONDITION
MAXIMA AND MINIMA
A second-order condition for f(x) to have a maximum at some value x = x0 is that together with dy/dx = f ￠(x) = 0, the second derivative (the derivative of the derivative) be negative, that is,
Consider again the p maximization example, which is also illustrated in Figure. Taking the second derivative of the p function yields

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Problem. A monopolist’s total revenue and total cost functions are
a. Determine the output level that will maximize profit p.
b. Determine maximum p.
c. Determine the price per unit at which the p-maximizing output is sold.

39. PARTIAL DERIVATIVES AND MULTIVARIATE OPTIMIZATION
Consider the following function:
the first-order conditions for a maximum or a minimum are given by

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The second-order conditions for a maximum are given by

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The second-order conditions for a minimum are given by:

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Example
Consider the following explicit relationship:

43. PRESENTATION 3

44. CONSTRAINED OPTIMIZATION
The manager often is required to maximize some objective function subject to one or more side constraints.
A production manager, for example, may be required to maximize the total output of a given commodity subject to a given budget constraint and fixed prices of factors of production.
Alternatively, the manager might be required to minimize the total costs of producing some specified level of output.

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The maximization or minimization problem can be written as:

46. Continue….. Example Minimize: TC (x, y) = 3x2 + 6y2 -xy
The total cost function of a firm that produces its product on two assembly lines is given as
TC (x, y) = 3x2 + 6y2 – xy
The problem facing the firm is to determine the least-cost combination of output on assembly lines x and y subject to the side condition that total output equal 20 units. This problem may be formally written as
Minimize: TC (x, y) = 3x2 + 6y2 -xy
Subject to: x + y = 20

47. SOLUTION METHODS TO CONSTRAINED OPTIMIZATION PROBLEMS
There are generally two methods of solving constrained optimization problems:
1. The substitution method
2. The Lagrange multiplier method

48. SUBSTITUTION METHOD
The substitution method involves first solving the constraint, say for x, and substituting the result into the original objective function. Consider, again, the foregoing example.
Minimize: TC (x, y) = 3x2 + 6y2 - xy
Subject to: x + y = 20

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In other words, this problem reduces to one of solving for one decision variable, y, and inserting the solution into the objective function.
Taking the first derivative of the objective function with respect to y and setting the result equal to zero, we get

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the second-order condition for total cost minimization is also satisfied:
Substituting y = 7 into the constraint yields
x + 7 = 20
x = 13
Finally, substituting the values of x and y into the original TC function yields:

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53. LAGRANGE MULTIPLIER METHOD
Sometimes the substitution method may not be feasible because of more than one side constraint, or because the objective function or side constraints are too complex for efficient solution.
Here, the Lagrange multiplier method can be used, which directly combines the objective function with the side constraint(s).

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The first step in applying the Lagrange multiplier technique is to first bring all terms to the right side of the equation.
20 - x - y = 0
With this, we can now form a new objective function called the Lagrange function, which will be used to find solution values to constrained optimization:

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Note that this expression is equal to the original objective function, since all we have done is add zero to it. That is, always equals f for values of x and y that satisfy g.
To solve for optimal values of x and y, we now take the first partials of this more complicated expression with respect to three unknowns—x, y, and .

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this system of equations may be solved simultaneously.The solution values are
x = 13; y = 17;l = -71

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it can be demonstrated that the Lagrange multiplier is defined as
That is, the Lagrange multiplier is the marginal change in the maximum value of the objective function with respect to parametric changes in the value of the constraint.

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In the context of the present example,l = -71 says that if we relax our production constraint by, say, one unit of output (i.e., if we reduce output from 20 units to 19 units), our total cost of production will decline by $71.

59. Problems
A profit-maximizing firm faces the following constrained maximization problem:
Determine profit-maximizing output levels of commodities x and y subject to the condition that total output equals 12 units.

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Find the first derivatives and the indicated values of the derivatives.
The total cost function of a firm is given by:
TC = Q Q2

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where TC denotes total cost and Q denotes the quantity produced per unit of time.
a. Graph the total cost function from Q = 0 to Q = 100.
b. Find the marginal cost function.
c. Find the marginal cost of production from Q = 0 to Q = 100.

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a. Determine for each equation the average variable cost, average cost, and marginal cost equations.
b. Plot each equation on a graph.
c. Use calculus to determine the minimum total cost for each equation.

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A firm has the following total revenue and total cost functions:
a. At what level of output does the firm maximize total revenue?
b. the firm’s total profit as p = TR - TC.
At what level of output does the firm maximize total profit?
c. How much is the firm’s total profit at its maximum?

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Assume that the firm’s operation is subject to the following production function and price data:
Q = 3X + 5Y – XY
Px = $3;Py = $6
where X and Y are two variable input factors employed in the production of Q.
a. In the unconstrained case, what levels of X and Y will maximize Q?

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b. It is possible to express the cost function associated with the use of X and Y in the production of Q as TC = 3X + 6Y. Assume that the firm has an operating budget of $250. Use the Lagrange multiplier technique to determine the optimal levels of X and Y.What is the firm’s total output at these levels of input usage?