The Calculus Primer (2011)

Part IV. Using the Derivative

Chapter 15. MAXIMA AND MINIMA

4—8. Changing Values of a Function. If a function increases as the independent variable increases, or decreases as the independent variable decreases, the function is said to be increasing; if the function decreases as the independent variable increases, or increases as the independent variable decreases, the function is decreasing. In other words, as we move left to right from a to 6, f(x) is increasing, or the curve is rising; as we move from b to c, the function is decreasing, or the curve is falling; from c to d, the function is again increasing.

We have already noted that a function is

(a) increasing when = f′(x) is positive (slope = tan ϕ is +),

(b) decreasing when = f′(x) is negative (slope = tan ϕ is − ).

It will be seen, therefore, in order that a function shall change from an increasing to a decreasing function, or vice versa, the sign of its first derivative must change from + to −, or from − to +. In other words, at some point the first derivative must have the value zero. Such a point is called an ordinary turning point, and is illustrated by points B, C, D, E, and I. At ordinary turning points such as these,

= f′(x) = 0,

the inclination of the tangent is zero, and the tangent at the turning point is parallel to the X-axis.

4—9. Maximum and Minimum Values of a Function. From the foregoing it will be seen that a function may have one or more maximum or minimum values. A maximum value of a function is one that is greater than any values immediately preceding or following; a minimum value of a function is one that is less than any values immediately preceding or following. Thus, in the figure, points B, D, G, and I are maximum points; points C, E, and H are minimum points.

Several important observations should be made.

(1)A function may have several maximum and minimum values.

(2)A maximum value of a function is not necessarily the greatest value the function may have, nor is a minimum value the least value. For example, there are values of y to the left of B which are greater than RB; there are values of y to the right of H which are less than SE and TE.

(3)At a maximum and minimum turning point (e.g., B, C, D, E, and I) the tangent is parallel to the X-axis, that is,

slope = = f′(x) = 0.

(4)At a point of inflection (e.g., A and F), the curve changes from concave upward to concave downward arc, or vice versa; the slope, f′ (x), does not change in sign, but the tangent passes through the curve.

(5)At special points (e.g., G and H), although the derivative may not exist, the tangent (or the curve) may be perpendicular to the X-axis, although this need not necessarily be the case.

It is apparent, therefore, that for a function to have a maximum or a minimum value, it is necessary for the value of f′(x) to be zero or infinite, but that this alone is not a sufficient condition. In addition to f′(x) having a zero or infinite value, f′(x) must change in sign as it passes through zero (or infinity).

We may then say that, in general, a function will have a maximum or minimum value under the following conditions:

(a)f(x) is a maximum if f′(x) = 0, and f′(x) changes from + to −.

(b)f(x) is a minimum if f′(x) = 0, and f′(x) changes from − to +.

Values of the independent variable at maximum and minimum points, or at exceptional turning points, are called critical values.

4—10. Rule for Finding Maxima and Minima. From the above discussion, we arrive at the following rule for determining the maximum and minimum values of a function:

Step 1.Find the first derivative of the function.

Step 2.Set the first derivative equal to zero and solve the resulting equation for real roots; these roots are critical values of the variable.

Step 3.Test the first derivative, using one critical value at a time, for a value first a little less, then a little greater, than the critical value. If the sign of the derivative changes from + to −, the function has a maximum value for that particular critical value of the variable; if the sign of the derivative changes from − to +, there is a minimum value; if the sign of the derivative does not change, then there is neither a maximum nor a minimum at that point.

EXAMPLE 1.Test for maximum and minimum values:

y = x² − 9x − 6.

Solution. = 2x − 9.

Put 2x − 9 = 0.

Solving, x = = 4 = critical value.

Testing near x = 4:

when x = 4, = −,

when x = 5, = +.

Hence the value x = 4 is a minimum value.

EXAMPLE 2.Test for maxima and minima:

y = x³ − 27x + 2.

Solution. = 3x² − 27 = 0.

3(x² − 9) = 0,

(x + 3)(x − 3) = 0,

x = −3, 3, critical values.

Testing x = −3;

Testing x = +3:

EXAMPLE 3.Divide the number k into two parts such that their product is a maximum.

Solution.Let one part be x; then the other part is (k − x).

ThenP = x(k − x),

P = kx − x².

= k − 2x.

Letk − 2x = 0,

then x = , the value of x which makes the product of x(k − x) a maximum.

EXAMPLE 4.Find the number which, when added to its square, yields a minimum sum.

Solution.Let x be the required number.

ThenS = x + x².

= 1 + 2x = 0.

x = −, the required number.

4—11. An Alternative Method for Determining Maxima and Minima. From the diagram it will be seen that in the vicinity of a maximum value of f(x), such as point R, as we pass along the graph from left to right, f′(x), the slope of f(x), changes from + to 0 to −. Thus f′(x) is a decreasing function; therefore, by §4—2 we know that its derivative (that is, f″(x), or the second derivative of the function itself) is negative or zero.

In the same way, in the vicinity of a minimum value of f(x), the slope of f(x), or f′(x), changes from − to 0 to +, and f′(x) is an increasing function; hence by §4—2, f″(x) is positive or zero.

At a maximum value, the curve is said to be concave downwards; at a minimum value, the curve is said to be concave upwards. These considerations lead to an alternative method of determining maxima and minima, based on the following principles:

(A)f(x) is a maximum if f′(x) = 0 and f″(x) is negative.

(B)f(x) is a minimum if f′(x) = 0 and f″(x) is positive.

Thus we see that if = 0 and is +, the curve is concave upward; if = 0 and is −, the curve is concave downward.

The rule of procedure is as follows:

Step 1. Find the first derivative of the function.

Step 2. Put the first derivative equal to zero and solve the equation thus obtained for real roots to determine the critical values of the variable.

Step 3. Find the second derivative of the function.

Step 4. Substitute each critical value for the variable in f″(x). If f″(x) is negative, there is a maximum value; if f″(x) is positive, there is a minimum value. If f″(x) = 0, the test fails; in such a case, there may or may not be a maximum or a minimum, and the method of §15—10 must be used.

EXAMPLE.Test the function y = x³ − 48x + 8 for maximum and minimum values.

Solution.f′(x) = 3x² − 48.

Put3x² − 48 = 0.

Hence x = 4, −4 are critical values.

f″(x) = 6x.

When x = 4, f″(x) = +; hence x = 4 is a minimum.

When x = −4, f″(x) = − ; hence x = −4 is a maximum.

EXERCISE 4—3

Test each of the following for maximum and minimum values:

1. y = x² − 4x;

2. y = x² + 6x + 8

3. y = x³ − 12x

4. y = x⁴ − 18x² + 15

5. y = x³ − 3x² + 6x − 2

Find the following:

6. Two numbers whose sum is 40 and whose product will be as large as possible.

7. A number which, diminished by its square, will be a maximum.

8. A number which, when added to its reciprocal, will give the smallest possible sum.

4—12. Applications of Maxima and Minima. Many practical situations arise in which the determination of maximum or minimum values is helpful.

EXAMPLE 1.A rectangle is to have a perimeter of 60 in.; find the dimensions which will give the maximum area.

Solution.A = lw.(1)

2l + 2w = 60,

l + w = 30,

l = 30 − w.(2)

Substituting (2) in (1):

A = w(30 − w) = 30w − w²,

= 30 − 2w.

For A to be a maximum, = 0.

Hence30 − 2w = 0,

w = 15.

From (2)l = 15.

Thus the rectangle, to have a maximum area, must be a square, 15 × 15.

NOTE. It is easy to prove that any rectangle of fixed perimeter, to have a maximum area, must be a square. This is left as an exercise for the reader.

EXAMPLE 2.A cylindrical tin can closed at both ends is to have a given fixed capacity, V. Prove that the amount of tin used (total surface of cylinder) will be a minimum when the height of the can equals the diameter of its base.

Solution.

Total surface = A = 2πr² + 2πrh;

Substituting for h:

4πr³ = 2V.

Substituting πr²h for V:

4πr³ = 2πr²h,

2r = h,

orh = 2r = diameter.

EXAMPLE 3.Find the altitude of a cylinder of maximum volume inscribed in a sphere of radius k.

Solution.OA =

h = AB = 2(OA) = 2.

V = πr²h = 2πr²,

Setting and solving:

3r² = 2k²,

r² = , for max. value of V.

andh = 2(OA) = k, for max. value of V.

EXAMPLE 4.An open box of greatest possible capacity is to be made from a square piece of cardboard whose sides are each 36 in. long, by cutting equal small squares out of the corners and folding up the remaining piece as suggested by the diagram. What should be the length of each side of the small squares?

Solution.Let x = side of small square; then 36 − 2x = side of square bottom of box.

V = x(36 − 2x)²,

= – 4x(36 – 2x)+(36–2x)²

= 12(x – 18) (x – 6).

Putting = 0: x = 18, x = 6; critical values.

If x = 18, there is no bottom, and the box has a minimum capacity;

If x = 6, the box is 24 × 24 × 6, and has a maximum capacity.

EXERCISE 4—4

1. An open trough is to be made from a long rectangular sheet of metal by bending up the long edges so as to give the trough a rectangular cross-section. If the width of the sheet is a inches, how deep should the trough be made so that it will have a maximum carrying capacity?

2. Solve illustrative Example 4 if the open box is made from a square a inches on a side.

3. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius a inches. Hint: r² + (h − a)² = a².

4. If the strength of a beam with rectangular cross-section varies directly as the width (w) and as the square of the height (h), what are the dimensions of the strongest beam that can be cut from a log of diameter d? Hint: Strength = S = kwh² is the function whose maximum we wish; S = kw(d² − w²).

5. Find the altitude of the right circular cylinder having a maximum lateral area that can be inscribed in a sphere of radius a. Hint:

6. Find the volume of the largest right circular cone that can be generated by revolving a right triangle of hypotenuse a inches about one of its sides.

7. A Norman window consists of a rectangle surmounted by a semicircle. For a given perimeter, find the height and width which will admit the maximum amount of light.

EXERCISE 4—5

Review

1. If a body moves in a fixed direction so that s = , prove that the acceleration is negative and proportional to the cube of the velocity.

2. At the end of t seconds, an object has a velocity of (2t² + 8t) ft. per sec.; find its acceleration (a) in general, and (b) at the end of 5 seconds.

3. Determine the maximum and minimum values of

y = x³ − 3x² − 24x + 10.

4. A circular filter paper of radius 9 inches is to be folded into a conical filter. Find the radius of the base of the filter if it has the maximum capacity.

5. The distance in feet covered in t seconds by a moving point is given by

s = 50t − 10t².

Find its velocity and acceleration at the end of 2 seconds.

6. Determine the maximum and minimum values of y = x³ − x² − 5x.

7. Determine the maximum and minimum values of y = x⁴ − 2x² + 3.

8. A freight train left the yards and in t hours was at a distance s = t³ − t² + 8t miles from the place where it started. Find its acceleration at the end of 1 hours.

9. If the sum of the length and girth of a cylindrical package must not exceed 90 inches, find the dimensions of the largest cylindrical package affording the largest volume and also meeting this requirement.

10. Find the area of the largest rectangle that can be inscribed in a circle of radius k.