In many cases we can find maximum or minimum values for a function just by studying the sign of its first derivative. Suppose f (x)is differentiable in an intervalI and that it has only one stationary point,x =c. Supposef(x)≥ 0 for allx inI such thatx ≤ c, whereasf(x)≤0 for allxinI such thatx ≥c. Thenf (x)is increasing to the left ofc and decreasing to the right ofc. It follows thatf (x)≤f (c)for allx ≤c, andf (c)≥f (x) for allx≥c. Hence,x =cis a maximum point forf inI, as illustrated in Fig. 1.
y ⫽ f(x)
c I y
x
y ⫽ f(x)
I d y
x Figure 1 x=cis a maximum point Figure 2 x=dis a minimum point With obvious modifications, a similar result holds for minimum points, as illustrated in Fig. 2. Briefly stated:1
1 Many books in mathematics for economists instruct students always to check so-called second- order conditions, even when this first-derivative test is much easier to use.
S E C T I O N 8 . 2 / S I M P L E T E S T S F O R E X T R E M E P O I N T S 263 T H E O R E M 8 . 2 . 1 ( F I R S T - D E R I V A T I V E T E S T F O R M A X I M U M / M I N I M U M )
Iff(x)≥0 forx ≤candf(x)≤0 forx ≥c, thenx =cis a maximum point forf.
Iff(x)≤0 forx ≤candf(x)≥0 forx ≥c, thenx =cis a minimum point forf.
E X A M P L E 1 Measured in milligrams per litre, the concentration of a drug in the bloodstreamt hours after injection is given by the formula
c(t )= t
t2+4, t ≥0 Find the time of maximum concentration.
Solution: Differentiating with respect tot yields c(t )= 1ã(t2+4)−tã2t
(t2+4)2 = 4−t2
(t2+4)2 =(2+t )(2−t ) (t2+4)2
Fort ≥ 0, the term 2−t alone determines the algebraic sign of the fraction, because the other terms are positive. In fact, ift ≤2, thenc(t )≥0, whereas ift≥2, thenc(t )≤0. We conclude thatt=2 maximizesc(t ). Thus, the concentration of the drug is highest 2 hours after injection. Becausec(2)=0.25, the maximum concentration is 0.25 milligrams.
E X A M P L E 2 Consider the functionf defined for allxby
f (x)=e2x−5ex+4=(ex−1)(ex−4) (a) Find the zeros off (x)and computef(x).
(b) Find the intervals wheref increases and decreases, and determine possible extreme points and values.
(c) Examine limx→−∞f (x). Sketch the graph off. Solution:
(a) f (x)=(ex −1)(ex−4)=0 whenex =1 and whenex =4. Hencef (x)=0 for x = 0 and forx = ln 4. By differentiating the first expression forf (x), we obtain f(x)=2e2x−5ex.
(b) f(x) =2e2x −5ex =ex(2ex −5). Thusf(x) = 0 forex = 5/2 = 2.5, that is, x =ln 2.5. Furthermore,f(x)≤0 forx ≤ln 2.5, andf(x)≥0 forx ≥ln 2.5. So f (x)is decreasing in the interval(−∞,ln 2.5] and increasing in [ln 2.5,∞). Hence f (x)has a minimum atx = ln 2.5, andf (ln 2.5) = (2.5−1)(2.5−4) = −2.25.
Sincef (x)→ ∞asx→ ∞,f (x)has no maximum.
(c) Whenx→ −∞, thenextends to 0, andf (x)tends to 4. The graph is drawn in Fig. 3.
(y=4 is an asymptote asx → −∞.)
⫺2
⫺3
⫺4 ⫺1 1
3 4
2 1
⫺1
⫺2 y
x
3 2 1
⫺1 1 2
y
x Figure 3 f (x)=e2x−5ex+4 Figure 4 f (x)=ex−1−x
Extreme Points for Concave and Convex Functions
Recall the definitions of concave and convex functions in Section 6.9. Suppose thatf is concave, withf(x)≤0 for allxin an intervalI. Thenf(x)is decreasing inI. Iff(c)=0 at an interior pointcofI, thenf(x)must be≥0 to the left ofc, whilef(x)≤0 to the right ofc. This implies that the function itself is increasing to the left ofcand decreasing to the right ofc. We conclude thatx=cis a maximum point forf inI. We obviously get a corresponding result for a minimum of a convex function.
T H E O R E M 8 . 2 . 2 ( M A X I M U M / M I N I M U M F O R C O N C A V E / C O N V E X F U N C T I O N S ) Supposef is a concave (convex) function in an intervalI. Ifcis a stationary point forf in the interior ofI, thencis a maximum (minimum) point forf inI.
E X A M P L E 3 Consider the functionf defined for allxbyf (x)=ex−1−x. Show thatf is convex and find its minimum point. Sketch the graph.
Solution: f(x)=ex−1−1 andf(x)=ex−1 >0, sof is convex. Note thatf(x)= ex−1−1=0 forx=1. From Theorem 8.2.2 it follows thatx =1 minimizesf. See Fig. 4 for the graph.
P R O B L E M S F O R S E C T I O N 8 . 2
1. Letydenote the weekly average quantity of pork produced in Chicago during 1948 (in millions of pounds) and letxbe the total weekly work effort (in thousands of hours). Nichols estimated the relation
y= −2.05+1.06x−0.04x2
Determine the value ofxthat maximizesyby studying the sign variation ofy.
S E C T I O N 8 . 2 / S I M P L E T E S T S F O R E X T R E M E P O I N T S 265
⊂SM⊃2. Find the derivative of the function hdefined for all x by the formula h(x) = 8x 3x2+4. Note thath(x)→0 asx→ ±∞. Use the sign variation ofh(x)to find the extreme points ofh(x).
3. The height of a flowering plant aftertmonths is given by h(t )=√
t−12t, t∈[0,3]
At what time is the plant at its tallest?
4. Show that
f (x)= 2x2
x4+1 ⇒ f(x)=4x(1+x2)(1+x)(1−x) (x4+1)2 and find the maximum value off on [0,∞).
5. Find possible extreme points forg(x)=x3lnx,x∈(0,∞).
6. Find possible extreme points forf (x)=e3x−6ex,x∈(−∞,∞).
7. Find the maximum ofy=x2e−xon [0,4].
⊂SM⊃8. Use Theorem 8.2.2 to find the values ofxthat maximize/minimize the functions given by the following formulas:
(a) y=ex+e−2x (b) y=9−(x−a)2−2(x−b)2 (c) y=lnx−5x, x >0 9. Considernnumbersa1,a2,. . .,an. Find the numberx¯which gives the best approximation to
these numbers, in the sense of minimizing
d(x)=(x−a1)2+(x−a2)2+ ã ã ã +(x−an)2
HARDER PROBLEM
⊂SM⊃10. After the North Sea flood catastrophe in 1953, the Dutch government initiated a project to determine the optimal height of the dykes. One of the models involved finding the value ofx minimizing
f (x)=I0+kx+Ae−αx (x≥0)
Herexdenotes the extra height in metres that should be added to the dykes,I0+kxis the con- struction cost, andAe−αxis an estimate of the expected loss caused by flooding. The parameters I0,k,A, andαare all positive constants.
(a) Suppose thatAα > kand findx0>0 that minimizesf (x).
(b) The constantAis defined asA=p0V (1+100/δ), wherep0is the probability that the dykes will be flooded if they are not rebuilt,Vis an estimate of the cost of flood damage, and δis an interest rate. Show thatx0= 1
αln αp0V
k
1+100 δ
. Examine what happens to x0when one of the variablesp0,V,δ, orkincreases. Comment on the reasonableness of the results.2
2 The problem is discussed in D. van Dantzig, “Economic Decision Problems for Flood Prevention”.
Econometrica, 24 (1956): 276–287.