Y = f (x) Y Y. x 0 x 1 x. Managerial Economics -- Some Mathematical Notes. Basic Concepts: derivatives.

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1 Managerial Economics -- Some Mathematical Notes I Basic Concepts: derivatives. The derivative of a function denotes its rate of change as a result of change in the right hand side variable(s). When we have me than one variable on the right hand side, this is then the "partial" derivative, in that it denotes only the change resulting from one right hand side variable changing, holding the others constant. Assume that Y=f(x). Then if we have a change in x, sax from x to x 1, the change in Y will be Y = f(x 1 ) - f(x ). We can also write the relationship between changes in Y and changes in X as follows: Y X = Y Y x x 1 1 Which simply gives us the rate of change in Y f a given change in x. This is shown graphically below: Y Y = f (x) Y 1 Y The slope of this line is dy/dx evaluated at the point x. Y x x 1 x x The derivative is simply the rate of change in Y following from a change in x, f very small changes in x. Hence, it is a measure of the slope at a point on the function. Our written expression f the derivative is dy/dx. When taking the derivative of any exponential function, we simply multiply the function by its exponent, and divide by the variable. F example:

2 If Q then dl = L 3 3 L L 3 L = F H G I K J = 3 Other rule f other functional fms (logs, trigonometry operats, etc.) can be found in a standard calculus text. Partial derivatives involve taking the derivative with respect to one variable in a multivariate equation. This is represented by the operat instead of d. Hence, if Q = a + b P c + d P e 1 Then we will have: Q = P b c P 1 1 c 1 Q = d e P P e 1 Notice that the derivative of any constant term is zero. II Basic Concepts: maximization and minimization: Consider the two functions below: Y,Z Minimum point, slope =, dyd/x = Y(x) Maximum point, slope =, dzd/x = Z(x) x

3 Notice that at a maximum minimum, the functions "flatten out". This means that, at the point of the maximum minimum, the slope is zero. Hence, since the derivative gives the slope, we can find this point by setting the derivative to zero. There is a problem, as illustrated above. We may have a maximum OR a minimum. We won't have both at the same time. How do we know which we have? Notice above that, at a minimum, the slope starts to rise as we move away. In other wds, the rate of change in the slope as we move away to the right from the minimum is positive. Hence, if take the derivative of the derivative, and look at its sign, we will know if we have a maximum minimum. If it is positive, we have a minimum, ese we have a maximum. F example, assume that R = a T b T c Then we can find the minimum/maximum point by setting the first derivative to zero: dr dt = a b c T c 1 = To find if it is a maximum minimum, we take the derivative of the derivative (called the second derivative): d dt F dri d R b c c T HG c dt K J = = ( 1) dt If this is positive, we have a minimum. If it is negative, we have a minimum III Example 1: Marginal product and output maximization Assume that we have the production function Q = L+. 5L. 5L 3 Questions: Find the marginal product of lab Find where output is maximized How does maximum output relate to the marginal product of lab? We can find the maximum/minimum point by setting the first derivative to zero:

4 = L. 15L = dl 1 L =. 15L, = L L = The relevant solution value f this quadratic equation is 6. (There are actually two roots, but the other is negative.) Is this a maximum a minimum? To find out, we need to take the value of the second derivative at the point L=6. d Q =. 3L dl L so that d = 6 Q 3, = <. dl This means we have a maximum. The function is shown below, where the maximum can be seen to be at L=6. The second derivative is also shown. In this case, it is the marginal product of lab. We can now see that When the marginal product is zero, total output with respect to that input is maximized. The marginal product will become negative (i.e. its rate of change, the second derivative of the production function) is negative as we move away from the maximum. L Q MPL Q MPL

5 IV Example : Maximization of revenue and profit Now, assume we have the following cost and demand functions: C( q) = 1. 5Q+. 5Q q = 1 P Questions: What level of output that maximizes revenue? What level of output that maximizes profit? What is the relationship of maximum revenue and profit to marginal revenue and marginal cost? What level of output minimizes average costs? What is the relationship of minimized average costs to marginal cost? From the demand and cost equations above we can derive the following relationships Inverse Demand (from rearranging demand): P = 5 Q / Revenue: R evenue = R = PQ = 5Q Q / Marginal Revenue: ma rginal revenue = MR = dr = 5. 1Q Marginal Cost: ma rginal c ost = MC = dc = Q Profit: P rofit = π = R C = (5 Q. 5Q ) ( 1. 5Q+. 5Q ) Lets consider maximization of revenue. From our discussion above, this will involve finding the point where MR=, as this denotes a maximum/minimum point. (It denotes a maximum here). Solving our MR function f zero involves the following: MR = 5. 1Q =, 5 =. 1Q, Q = 5. Hence, revenue is maximized at 5. This is illustrated below:

6 Revenue Next lets consider profit. Maximizing profit means setting the first derivative of the profit function to zero, per our discussion above. Remember that profit involves R and C (revenue and cost). This means that the first derivative will involve dr and dc. Fmally, P rofit = π = R C = (5 Q. 5Q ) ( 1. 5Q+. 5Q ) sothat dπ dr dc = = MR MC = rearranging, we then have MR = MC 5. 1Q = Q, so that 5. 5 =. Q, Q = 5. 5 Profit is apparently maximized ( perhaps minimized) at Q=5.5. We can show this is a maximum as follows: dπ = MR MC = 55.. Q so that d π =. < ( wehave a maximum)

7 This is shown in the table and figure below: Quantity Total Cost Average Cost Revenue Profit Marginal Revenue Marginal cost Profit Marginal Revenue Marginal cost Demand

8 This example illustrates the following points: MR= at the point where revenue is maximized MR=MC at the point where profit is maximized The MR curve cuts below the demand curve As MR turns negative, total revenue starts to fall Lets now turn to the cost side of our example. We do in fact have scale economies f small ranges of output, as illustrated in the table below: Quantity Total Cost Average Total Cost Marginal Cost

9 Relating the table to the calculus, we have from befe that ma rginal c ost = MC = dc = Q This will be zero at the cost minimum. Hence MC = Q =, so that. 1Q =. 5, Q = 5 So that total costs are minimized at 5 units of output. Consider next average cost. This involves dividing total cost by Q. c( Q) = 1. 5Q+. 5Q AC c ( Q ) 1 = = Q Q Q Minimizing AC involves setting dac/ =. We then have 1 AC = 1Q Q so that from our rule above on ex ponential derivatives dac Q Q ( 1) ( 1) 1 1 = ( 1) 1 Q Q = 1Q +. 5 = 1 = = F H G I K J = Average costs are minimized at Q=44.7. We have graphed this below. MC cuts AC at the minimum. 5 Average Total Cost Marginal Cost