1.3. ApplicationsMaxima and minimaWhen a function goes up and then smoothly turns around and comes back down again, it has zero slope at the top. A place where \(\dot{x}=0\), then, could represent a place where x was at a maximum. On the other hand, it could be concave up, in which case we'd have a minimum. The term extremum refers to either a maximum or a minimum. Example 7◊ Fred receives a mysterious email tip telling him that his investment in a certain stock will have a value given by x = 2t^{4} + (6.4577×10^{10})t, where \(t\ge 2005\) is the year. Should he sell at some point? If so, when? ◊ If the value reaches a maximum at some time, then the derivative should be zero then. Taking the derivative and setting it equal to zero, we have \begin{align} & 0 = 8t^3+6.4577\times10^{10} \\ & t = \left(\frac{6.4577\times10^{10}}{8}\right)^{1/3} \\ & t = \pm 2006.0. \end{align} Obviously the solution at t = 2006.0 is bogus, since the stock market didn't exist four thousand years ago, and the tip only claimed the function would be valid for \(t\ge 2005\). Should Fred sell on New Year's eve of 2006? But this could be a maximum, a minimum, or an inflection point. Fred definitely does not want to sell at t = 2006 if it's a minimum! To check which of the three possibilities hold, Fred takes the second derivative: \[\ddot x = 24t^2 .\] Plugging in t = 2006.0, we find that the second derivative is negative at that time, so it is indeed a maximum. Implicit in this whole discussion was the assumption that the maximum or minimum occurred where the function was smooth. There are some other possibilities. In Figure N, the function's minimum occurs at an endpoint of its domain. Another possibility is that the function can have a minimum or maximum at some point where its derivative isn't well defined. Figure o shows such a situation. There is a kink in the function at t=0, so a wide variety of lines could be placed through the graph there, all with different slopes and all staying on one side of the graph. There is no uniquely defined tangent line, so the derivative is undefined. Example 8◊ Rancher Rick has a length of cyclone fence L with which to enclose a rectangular pasture. Show that he can enclose the greatest possible area by forming a square with sides of length L/4. ◊ If the width and length of the rectangle are t and u, and Rick is going to use up all his fencing material, then the perimeter of the rectangle, 2t+2u, equals L, so for a given width, t, the length is u = L/2  t. The area is a = tu = t(L/2  t). The function only means anything realistic for 0≤ t≤ L/2, since for values of t outside this region either the width or the height of the rectangle would be negative. The function a(t) could therefore have a maximum either at a place where \(\dot{a}=0\), or at the endpoints of the function's domain. We can eliminate the latter possibility, because the area is zero at the endpoints. To evaluate the derivative, we first need to reexpress a as a polynomial: \[ a=t^2+\frac{L}{2}t . \]
The derivative is \[ \dot{a}=2t+\frac{L}{2} . \]
Setting this equal to zero, we find t = L/4, as claimed. This is a maximum, not a minimum or an inflection point, because the second derivative is the constant \(\ddot{a}=2\), which is negative for all t, including t = L/4. Propagation of errorsThe Women's National Basketball Association says that balls used in its games should have a radius of 11.6 cm, with an allowable range of error of plus or minus 0.1 cm (one millimeter). How accurately can we determine the ball's volume? Figure P. How accurately can we determine the ball's volume? The equation for the volume of a sphere gives V=(4/3)π r^{3}=6538 cm^{3} (about six and a half liters). We have a function V(r), and we want to know how much of an effect will be produced on the function's output V if its input r is changed by a certain small amount. Since the amount by which r can be changed is small compared to r, it's reasonable to take the tangent line as an approximation to the actual graph. The slope of the tangent line is the derivative of V, which is 4π r^{2}. (This is the ball's surface area.) Setting (slope)=(rise)/(run) and solving for the rise, which represents the change in V, we find that it could be off by as much as (4π r^{2})(0.1 cm)=170 cm^{3}. The volume of the ball can therefore be expressed as 6500±170 cm^{3}, where the original figure of 6538 has been rounded off to the nearest hundred in order to avoid creating the impression that the 3 and the 8 actually mean anything  they clearly don't, since the possible error is out in the hundreds' place. This calculation is an example of a very common situation that occurs in the sciences, and even in everyday life, in which we base a calculation on a number that has some range of uncertainty in it, causing a corresponding range of uncertainty in the final result. This is called propagation of errors. The idea is that the derivative expresses how sensitive the function's output is to its input. The example of the basketball could also have been handled without calculus, simply by recalculating the volume using a radius that was raised from 11.6 to 11.7 cm, and finding the difference between the two volumes. Understanding it in terms of calculus, however, gives us a different way of getting at the same ideas, and often allows us to understand more deeply what's going on. For example, we noticed in passing that the derivative of the volume was simply the surface area of the ball, which provides a nice geometric visualization. We can imagine inflating the ball so that its radius is increased by a millimeter. The amount of added volume equals the surface area of the ball multiplied by one millimeter, just as the amount of volume added to the world's oceans by global warming equals the oceans' surface area multiplied by the added depth. For an example of an insight that we would have missed if we hadn't applied calculus, consider how much error is incurred in the measurement of the width of a book if the ruler is placed on the book at a slightly incorrect angle, so that it doesn't form an angle of exactly 90 degrees with spine. The measurement has its minimum (and correct) value if the ruler is placed at exactly 90 degrees. Since the function has a minimum at this angle, its derivative is zero. That means that we expect essentially no error in the measurement if the ruler's angle is just a tiny bit off. This gives us the insight that it's not worth fiddling excessively over the angle in this measurement. Other sources of error will be more important. For example, is the book a uniform rectangle? Are we using the worn end of the ruler as its zero, rather than letting the ruler hang over both sides of the book and subtracting the two measurements?
