CHAPTER XII
MAXIMA AND MINIMA
if. In Chapter ix, para. 13, a definition was given of a function f (x) which is continuous at the point x = a. A property of such a function which is of frequent application in the calculus is as follows :
If f (x) is a continuous function throughout the range of values considered, then for values of x near the point x = a, f (x) has the same sign as f (a), provided that f (a) is not zero.
This proposition is almost self-evident. Since, for a continuous function, Lt f (x) is equal to f (a), it follows from the definition of
x-ka
a limit that there is a range of values over which f (x) differs from f (a) by less than E where E may be as small as we please. For any value of E numerically less than f (a) the sign of f (x) for values of x within the corresponding range will be the same as that of f (a).
Now let y = f (x) be a continuous function of x.
dy Ay
Then — Lt -
dx ,,,,. 0 Ax
so that Ox = dy
+ e where E is a small quantity whose limit as Ax -~ o is zero.
If dt is not zero, the sign of dx will be the same as that of 0~ provided that we take Ax small enough ; if Ax -> o, the sign of Ay will be the same as that of Ax dx.
Consequently if Ax is positive but -~ o, Ay will have the same d
sign as x
But Ay = f (x + Ax) — f (x).
Therefore f (x + Ax) — f (x) will have the same sign as dx if Ax is positive, but — o.
MAXIMA AND MINIMA 191
If Ax is positive, x + Lx is greater than x; i.e. x is increasing: and if dx is positive f (x + Ox) is greater than f (x) ; i.e. y is in-creasing.
Therefore f (x) increases as x increases if /x is positive, and decreases as x increases if dx is negative.
Similarly, if Ox is negative, so that x is decreasing, f (x) decreases as x decreases if is positive, and increases as x decreases if dx
is negative.
Values off (x) at which the function ceases to increase (decrease) and begins to decrease (increase) are called turning values or critical values.
2. Maxima and Minima.
At the points where the function y = f (x) ceases to increase and begins to decrease y is generally said to have a maximum value : conversely where the function ceases to decrease and begins to increase y is said to have a minimum value.
It should be noted that a maximum value need not necessarily
be the greatest numerical value of the function, nor need a minimum Y
value be the least. For example, in Fig. z6, there are maxima at the points A and C, and minima at B and D. The numerical value, however, of the ordinate at D is greater than that of the ordinate
at A, although the function assumes p X
a minimum value at D and a maxi- Fig. 26.
mum value at A.
The following is a more correct definition of maximum and minimum values :
The function y = f (x) has a maximum value at the point x = a, if f (a) exceeds both f (a + h) and f (a — h) for all positive values of h less than a small finite quantity E. Similarly, f (x) has a
192 DIFFERENTIAL CALCULUS
minimum value when x = a, if f (a) is less than both f (a + h) and f (a — h) for all positive values of h less than E.
If therefore f (a) is a maximum value of y = f (x), (i) as x in-creases from (a — h) to a, y increases and dx is positive ; and (ii) as x increases from a to (a + h), y decreases and dx is negative
(para. 1). That is, as x increases changes from a positive to a negative value. The criterion for a maximum value at x = a is therefore that dx changes sign from positive to negative as x passes
through a. Conversely, for a minimum value dx changes from negative to positive.
Since a continuous function cannot change sign without passing through a zero value, we have that, for a critical value, dx must be zero provided that it be continuous.
If therefore f (a) is a maximum or a minimum value of the function y = f (x), and f' (x) is continuous,
- (i)dymust be zero; [dx x=a
- (ii)dx must change from positive to negative for a maximum value ;
dx must change from negative to positive for a minimum value.Example 1.Find the maximum and minimum values ofy = 4x3 — 18x2 + 24x + II.(dx = 12x2 — 36X + 24.We must equate dy to zero : this gives
12x2 — 36X + 24 = 0,
MAXIMA AND MINIMA 193 x2 — 3X + 2 = 0,
x=2 or I.
These values of x give critical values to y.
To find which of these values gives a maximum and which a minimum we must proceed further.
Let x = 2 — Ox and 2 + Lx in turn, where Ax is a small positive quantity.
(i) dx = (2 — 0x)2 — 3 (2 — Ox) + 2, when x = 2 — Ox, =4—44x+(0x)2—6+3AX+2=—Ox+(0x)2;
- (ii)d- -x — 4 + 4AX + (0x)2 — 6 — 3AX + 2 =,x + (0x)2,
when x = 2 + Ox.Now since Ax is a small positive quantity,[— Ox + (0x)2] is negativeand[Lx + (0x)2] is positive.: , dx passes from negative to positive as x passes through the value 2. ... x = 2 gives y a minimum value.Similarly it may be shown that x = I gives y a maximum value. The values required are therefore
Maximum: y=4.13—18.12+24.I+II=21.
Minimum: y = 4 .23 — 18 .22 + 24 .2 + I I = 19.
3. An alternative method for determining the maximum or mini
mum values of a continuous function depends upon the rate of
change of dx. We have seen that if f (a) is a maximum value, dy
changes from positive as x passes through a. In other words d x is
decreasing near the point, and consequently its differential co
efficient, i.e. dx , or f" (x), must be negative. Therefore, by the
proposition in para. I, the sign off" (x) is the same as that off" (a)
provided that f" (a) is not zero.d2y
We have therefore for a maximum value at the point x = a,dxmust be negative : and conversely for a be positive.B132 minimum value dx2 must
194 DIFFERENTIAL CALCULUS
The test is easy to apply. For example, using the same function as in Ex. 1, we find that
dx -~-~ 12 (x2—3x+2) = 12(2x-3).
This is positive when x = 2 and negative when x = 1. Consex = 2 gives a minimum value and x = 1 a maximum (as above).
4. The tests for maximum and minimum values are quite straight-forward and their application to simple problems presents little difficulty. The following examples are illustrative of the methods employed.
Example 2.
A window is in shape a rectangle with a semicircle covering the top. If the perimeter of the window be a fixed length p, find its maximum area.
We have first to choose an independent variable. Let BO, the radius of the semicircle, be x. Then since the perimeter of the figure is a fixed length p,
p=2BC+CD+AKB
= 2BC + 2x + vrx,
so that BC=p—(2+7r)x
2
The area a will be [rectangle ABCD + semicircle AKB],
i.e. a = BC. CD + ~~rx2
p—(2+~r)x
. 2x + Z7rx2 2
=xp—(2+?r)x2+21rx2.
da dx=p-2x(2+7r)+7rx.
For a maximum or minimum value dx = o, i.e. x = p
4+ITIt is evident that this will give a maximum value to a, for when x = o
the area is zero. We need not therefore apply the second test.
Fig. 27.
ILLUSTRATIVE EXAMPLES 195 Giving x the above value,
a=xp — (2+7r)x2+ 17rx2
p2 _ 4 + ,T p2 p2
4+7r 2 (4+-ff)2 2(4+7r). Example 3.
Given u_I = — 5 ; ul = — I ; u2 = 4; U5 = 175 ; find the maximum and minimum values of ux.
Since four values of ux are given we must assume that the function is a rational integral function of the third degree in x.
| Let |
y=ux=a+bx+cx2+dx3. |
| Then |
-5=u_1=a—b+c—d,
—I=u1=a+b+c+d, |
| |
4=u2=a+2b+4c+8d, |
15 = u5 = a + 5b + 25c + 125d. Solving these equations we obtain easily that a=o; b=o; c=—3; d=2.
y=—3x2+2x3.
For critical values d= o,
x
i.e. — 6x + 6x2 = o,
giving x = o or 1.
Also d y
= — 6 + 12x.
When x = o, d a is negative, giving a maximum value;
x = 1, d z is positive, giving a minimum value.
Therefore maximum value of y is o ;
minimum value of y is — I.
Example 4.
A ladder is to he carried in a horizontal position round a corner formed by two streets a feet and b feet wide meeting at right angles. Prove that the length of the longest ladder that will pass round the corner without jamming is (a* + b*)* feet.
In this example it is advisable to take as the variable the angle that the ladder makes with the wall of the street. Call this angle a. Let x be
13-2
196 DIFFERENTIAL CALCULUS
the distance across the corner. The longest ladder will of course be the shortest distance across the corner, and the problem reduces to one of finding the minimum value of x. Then
x=AB=AC+CB
= a sec a + b cosec a,
dx_asecatana — bcosecacota; da
i.e. for a maximum or minimum value, a sec a tan a — bcosecacota=o, sin a cos a _
Or a cost a— b sing a — 0'
from which tan a = . a
B Fig. z8.
This evidently gives a minimum value to a sec a + b cosec a, the maxi-mum value being oo .
b* al
sina=(al+bl)I; Cosa=(al+b)I,
a b
and x = +
cos a sin a
= (al + b1)I
on simplifying.
5. Points of inflexion.
It has been seen above that, for a critical value at x = a, f' (a) = o and that in order to ascertain whether this value is a maximum or
a minimum, recourse must be had to the change of sign of f" (a), provided that f" (a) is not zero. The question of what happens if f" (a) is in fact zero now arises. Now f" (a) will be zero if there is no rate of change off' (a). In that event the first differential coeffiwill not increase (decrease) and then decrease (increase) as x passes through a, although
f'(x) = o for the value x = a. f'(x) will have the same sign for
X Fig. zg.
POINTS OF INFLEXION 197
the value x = a + h as for x = a — h where h is small. There is therefore, as a rule, no maximum or minimum value at the point A, and A is said to be a point of inflexion on the curve.
In general there is a point of inflexion at x = a if f" (a) is zero : the exceptions depend upon the values of the higher derivatives.
Example 5.
Find the points of inflexion on the curve y (1 + x3) = 1. =1+x3'
dv — 3x2
.. dx = (I + x3)2.
Therefore for a critical value, x = o. But d2y 2X
6x4
doe. - _ - 3 `(I + x3)2 — (1 + x3)3
which is zero when x = o or x = ,Z/1,7. There are therefore points of inflexion where x = o and 4 .
6. We have illustrated the critical values of the function y = f (x) by reference to the geometry of the curve. The problem may also be considered analytically.
By the extension of the Mean Value Theorem (p. 177)
f (a + h) = f (a) + hf' (a) + Zh2f" (a + Ojh), and f (a — h) = f (a) — hf' (a) + 1--h2f" (a — 02h),
where 01, B2 are positive proper fractions not necessarily equal. Now for critical values f' (a) = o.
Therefore f (a + h) — f (a) = +h2f" (a + 0,h), and f (a — h) — f (a) = 2h2 f" (a — 02h).
If h be made sufficiently small the right-hand side will have the same sign in both expressions. For a maximum value
f (a + h) — .f (a) and f (a — h) — f (a)
will both be negative. Therefore, since 2h2 is positive, the second differential coefficient must be negative. Similarly for a minimum
198 DIFFERENTIAL CALCULUS
value the second differential coefficient must be positive. If, however, f" (a) is zero we must consider a further term : thus f (a + h) = f (a) + hf' (a)+2Zi f„ (a)+-i fm (a + 0,h),
h' h'
.f (a — h) = f (a) — hf' (a) + 2 I f" (a) — -i f'" (a — B4h),
which reduce to
.f (a + h) — f (a) = 3 f,,, (a + 03h), 3
and f(a—h)—f(a)=—3f"' (a—04h).
It will be seen that here there can be no maximum or minimum value if f"' (a) is not zero, for since the sign of the right-hand side
h'
can be made to depend upon the sign of 3 ~ f"' (a), the signs of
f (a + h) — f (a) and f (a — h) — f (a) will be different and there will be a point of inflexion.
We may carry this proof further. If f"' (a) is zero, the condition for maxima and minima will depend upon the sign of f iv (a)—provided that f iv (a) is not zero—and so on. In general, therefore, we may say that
For a maximum or minimum value the first derivative that does not vanish must be of an even order : if in that event the derivative is negative the critical value is a maximum and if positive a mini-mum. Otherwise there will be a point of inflexion.
It is worthy of note that if f (x) is a continuous function, maximum and minimum values (if any) occur alternately.
Example 6.
Examine the critical values of y = x2 — 3x3 + 3x4 — x5. y=x2—3x3+3x4— x5.
- = 2X — 9x2 + 12x3 — 5X4.
Equating this to zero we obtain the four values x = o, 1, 1, . dx2y = 2 — 18x + 36x2 — 2ox3i
MISCELLANEOUS APPLICATIONS 199
— •72, giving a maximum value.
d3y
= — 18 + 72X — 6ox2;
dx3
d3y 6,
dx3
and since this is not zero x = 1 gives a point of inflexion.
- In the above demonstrations it has been assumed that the functions concerned have a differ
coefficient for all values of the variable considered. There are, how-ever, continuous functions which do not have a definite derivative forevery value of the variable, although X,pxthey may have a maximum or mini-};mum value at some point for whichFig. 30.there is no definite differential coefficient.For example, there is a minimum value at the point x = o on the curvey = x',although there is no definite derivative at that point. When x = o, x is infinite.
- We will conclude this chapter with some miscellaneous appliof the above processes. Example 7.
Find the maximum value of (x — a)2 (x — b).y = (x — a)2 (x — b),dywhen x = o,x = 1,x=!,Againwhen x = 1,d' -vdx2 = 2, giving a minimum value;d2ydx2 d2vdx2o, which must be examined further;dx=2(x—a) (x—b)+(x—a)2=(x—a)(3x—zb—a).If dx=o,then x=a, or a+2b; 3
200 DIFFERENTIAL CALCULUS
d 'y
=3x—2b—a+3(x—a)=6x—2b—4a.
If x=a, d
d2y
x =2a—2b, andifx=a 32b, d2=2b— 2a.
We cannot therefore determine the sign of the second differential coefficient unless we know the relative magnitudes of a and b.
If a > b, x = a + 2b gives y a maximum value,
3 a<b, x=a
a = b, there is a point of inflexion for the value x = a.
The required values of (x — a)2 (x — b) will be found by substitution of the values of x in the original expression.
Example S.
Divide the number 21 into three parts a, b, c in continued proportion such that 3a + 6b + 4C may be a maximum.
Since a, b, c are in continued proportion, a : b :: b : c, so that we may write a = bk and b = ck.
.'. a = ck2. But a+b+c=zl.
:. (I + k + k2) c = 21 (i).
We have to find the maximum value of 3a + 6b + 4C or, in terms of c and k, of (4 + 6k + 3k2) c.
Let (4 + 6k + 3k2) c = z
(ii).
Then if there were one variable k, the necessary procedure would be to find the values of k which give dk a zero value. But c may vary as well as k, and if we differentiate z with respect to k a further differential coefficient, namely dk , is involved. We can, however, make use of equation (i) and thus eliminate dk from our differential equations.
Differentiating equations (i) and (ii) respectively with respect to k,
dimc
(6k+6) c + (4 + 6k + dc_dz_0
3k,) dk dk
for a critical value.
ILLUSTRATIVE EXAMPLES 201
Eliminating dk we obtain easily that
3k2+2k—2=o,
so that k = — the positive root giving a maximum value. 3
Substituting in (i) and simplifying,
c=14-V7,
whence a = 14 — 4 'V7,
b=5'V7—7,
and the required value of 3a + 6b + 4C becomes 56 + 14 V7.
(The minimum value, found from the value k = — 'V5, is
3
56 — 14 V/7). Example 9.
Find the maximum value of 2 (a — x) (x + Vx2 + b2) where x is real.
In certain circumstances it may happen that a simple and straight-forward method of obtaining maximum or minimum values can be evolved by reference to algebra or geometry.
For example, although by differentiating the above expression and equating the result to zero the required value can be obtained, a neater proof results from the use of a well-known algebraic property.
(a — x)2 and x2 + b2 are positive since x is real.
| |
(a — x)2 + (x2 + b2) |
2 '(a — x)2 (x2 + b2), |
| i.e. |
|
2 (a — x) 1'x2 + b2, |
| i.e. |
a2 — tax + x2 + x2 + b2 |
2 (a — x) 1'x2 + b2, |
| or |
— 2 (a — x) x + a2 + b2 |
2 (a — x) 1/'x2+b'2, |
| i.e. |
a2 + b2 |
2 (a — x) (x +V x2 + b2). |
In other words, 2 (a — x) (x + 1'x2 + b2) is not greater than a2 + b2, i.e. the maximum value of the expression is a2 + b2.
202 DIFFERENTIAL CALCULUS
EXAMPLES 12
Find the maximum and minimum values, where such exist, of
7. 12x5 — 45x4 + 40x3 + 6. 9. x3—5x2+8x—4.
x3
2. — + ax2 — 3a2x.
3
4. x2 (I + x2)-3.
x2 — I
6. (3 + x2)3.
8 (x + a) (x + b) (x—a)(x—b)'
10. ax + by, where xy = c2.
- If ux = a2 + n2x2, find the value of x which gives u its smallest value.
Find the maxima and minima of 1 + z sin 9 + 3 cos2 B.If xy = 720 find to one place of decimals the minimum value of 5x + 3y.14. Find the maximum value of x. x
- Show that x3 — 3x2 + 6x + 3 has neither a maximum nor a mini-mum value.
Find the minimum value of 9X — 6x loge 3.Find the values of a and P in order that xs + ax5 +,8x4 may have a maximum value when x = 2 and a minimum value when x = 3.
- ABCD is a rectangular field; AB is 200 yards, BC is loo yards. A man has to walk from A to C. He can walk at 5 miles an hour down the side AB, but directly he leaves the path AB and strikes across the grass he can only go at 3 miles an hour. Find which is his quickest route.
An open box is to be made on a square base with vertical sides out of a given quantity of cardboard of area c2. What is the maximum volume of the box?Into how many parts must the number ne be divided so that their continued product may be a maximum; n being a positive integer and e the base of the Napierian logarithms?A rectangular piece of cardboard, sides a, b, has an equal square cut out of each corner. Find the side of the square so that the remainder may form a box of maximum volume.
EXAMPLES 203
- Find the length of the shortest straight line which can be drawn through the point (a, b) terminated by the rectilinear axes.
- A man is 2 miles from the nearest point A of a straight road, and he wishes to reach a point B on the road 4 miles from A. He can walk at 4 miles per hour until he reaches the road and at 5 miles per hour on the road. Find the least time in which he can reach B.
- Find the maximum and minimum values of y regarded as a function of the variable x, where ax2 + 2hxy + by2 + 2cx = o.
A fixed point is taken on a circle of radius r, and a chord is drawn from this point to any other point on the circle. The tangent to the circle at the second point is constructed and a perpendicular is dropped from the fixed point on to the tangent. Prove that the maximum area of the triangle formed by the chord, the tangent and the perpendicular is 3 \/3r2/8.In a submarine telegraph cable the speed of signalling varies asx2 log I , where x is the ratio of the radius of the core to that of the xcovering. Show that the greatest speed is attained when this ratio is r.el.
- A person being in a boat a miles from the nearest point A of the beach wishes to reach as quickly as possible a point B which is b miles from A along the shore. The ratio of his rate of walking to his rate of rowing is A. Find the distance from A at which he should land.
A wire of given length is cut into two portions which are bent into the shapes of a circle and a square respectively. Show that if the sum of the areas be the least possible the side of the square is twice the radius of the circle.An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expense of lining it with lead will be least if the depth is made half the width.
- Find the least value of ae.x + be-"x.
- Find the maximum area of the rectangle which can be drawn with its sides passing through the four corners of a given rectangle whose sides are a and b in length respectively.
- A train passes a station X at a rate of 30 miles per hour. Its speed increases and at any point exceeds the speed at X by a quantity proportional to the time elapsed since leaving X. At the end of a minute it passes Y, 3840 feet from X. A second train passes X 8 seconds after
Illustrate your results by drawing a graph of the function.
36. If x be the independent variable find the maximum and minimum values of y given
y — 12 — x4 (5x2 + 6x — 15) = 0.
37. Explain how to discriminate between the maxima and minima values of f (x), if
df
and a, b, c be in ascending order of magnitude.
- 38.Explain what is meant by a "point of inflexion" on a curve and show how to find the points of inflexion, if any, on the curve y = f (x). Find the points of inflexion on the curve
2y2 x2 (a2 — x2). a2
- 39.Draw a graph of the curve y = e-x2, and find the points of in
flexion.2
- 40.Find the points of inflexion on the curve y =x , and illustrate
1+xby a diagram.
- Prove that the triangle of maximum area inscribed in a circle is such that the tangents to the circle at the angular points are parallel to the opposite sides.
x2 2
- 42.If a + b = 1, show that the maximum and minimum values of
x2 + xy + y2 are the roots of the equation 4z2-4z(a+b)+3ab=o.
- 9y2 + 6xy + 4x2 — 24y — 8x + 4 = 0. Find the maximum and minimum values of y.
204DIFFERENTIAL CALCULUSthe first and travels at a uniform speed of 45 miles per hour. Find the minimum distance between the two trains at any time.
- A function y is the sum of two functions of which the first varies as the cube of x and the second inversely as the square of x. The least value of y is 5 which occurs when x = 1. Find the complete expression of y as a function of x.
x2
- Trace the curve y = x2 + +
+ 1 , finding the maximum and minimum values of y.
- Find the maximum and minimum values of y = x +.
X + 2
EXAMPLES 205
- 44.Given
log10 e = '4343; logic, 2 = '3oio; log10 3 = '4771, find the maximum and minimum values of12 (loge x + i) + x2 — Iox.45. Draw a rough sketch of a curvey (2x2 + 13X — 7) = iox2 + 30X, and find the maximum and minimum values of y.
- 46.If x2 + y2 = i, find the minimum value of 3x + 4y.
47. Find the minimum value ofa2 cos2 x + b2 sin2 xsin2 x cos2 x