Calculus and Optimization | Business Risk Management

As we will see in the lecture this coming Thursday, August 23rd, a?particularly important part of the analytic toolkit for the risk management course is?basic calculus.

Algebra and calculus share in common the problem of?calculating slope values.? The primary difference is that the algebra indicates slope values for discrete changes, whereas the calculus indicates slope values for continuous changes.? By a discrete change, I mean a change that is ?countable?; e.g., in the equation y = 10 + 5x, a discrete change (Dy) occurs?in y whenever a discrete change (Dx) occurs?in?x. For example, suppose?the initial value for x is?2 and then we change x?s value to 4.??When Dx = 2, ?y changes from?20 to 30, so Dy = 10 and Dy/Dx =?10/2 = 5.?By a continuous change, I mean a change that is infinitesimally small; specifically,?we study how an infinitesimally small?Dx (called dx) causes an infinitesimally small?Dy (called dy) to occur.? It turns out that in the case of our line example, dy/dx = Dy/Dx = 5.? In cases involving nonlinear functions such as a parabola (where y = x²), Dy/Dx is at best a crude approximation for dy/dx. However, as Dx ?> 0, Dy/Dx ?> dy/dx; in other words, the algebra result converges upon the calculus result.

In calculus, slope values are determined via a procedure called??differentiation?, and the slope?of a function such as a line or a curve is referred to as its ?first derivative?.??The first derivative indicates the rate at which one?variable (let?s call it y) changes with respect to?small changes in?another variable (let?s call the other variable x).? The second derivative?corresponds to the rate of change in the?slope itself.? The second derivative comes in handy whenever we try to maximize or minimize a function.? For example, suppose you are interested in determining how many units of a product to produce.??If you produce and sell Q units of this product at a price of P dollars per unit, then total revenue of TR = PQ.? If your total costs (TC) are fixed, then you would want to?maximize total revenue, since this would maximize profit.? However,?one also typically incurs?variable costs which increase as more units are produced, so in order to maximize profit, you?ll want to produce up to the point at which the revenue generated from selling the last unit of product (also known as marginal revenue, or MR) is equal to the cost incurred from producing that unit (also known as marginal cost, or MC).? Since marginal profit (MP) = MR?- MC = 0, this implies that total profit is either maximized or minimized when?MR = MC.

Suppose total profit is p = TR ? TC, where TR = 30Q and TC = 40?+ 3Q²; then?marginal profit?MP =?dp/dQ = dTR/dQ?- dTC/dQ?= 30 ??6Q, which equals 0 if Q = 5.? We know that this is maximum profit because the 2^nd derivative (which corresponds to?the slope of the slope of the total profit equation) is negative; i.e., d²p/dQ² = dMP/dQ = ?6 < 0.? Conceptually, a negative second derivative implies that as one moves away from Q = 5; e.g., by selecting either Q = 4 or Q = 6, then profit must be lower than at?Q = 5.? Note that at Q = 5, p = 30(5)?? 40 ? 3(5²)= 150 ? 115 = $35.? Profit at Q = 4 is only p = 30(4)?? 40 ? 3(4²)= 120 ??88 = $32, so producing 5 instead of 4 units results in $3 more profit.? Similarly, profit is also $3 lower at Q =?6 compared with Q = 5; note that if Q =?6, then p = 30(6)?? 40 ? 3(6²)= 180 ??148 = $32. Finally, also note that at Q = 5, MR = 30 and MC = 6(5) = 30!

Source: http://risk.garven.com/2012/08/21/calculus-and-optimization-2/

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