Maximum/Minimum Problems Help

Example 1

Find all local maxima and minima of the function k(x) = x ³ − 3 x ² − 24 x + 5.

Solution 1

We begin by calculating the first derivative:

k′ ( x ) = 3 x ² − 6 x − 24 = 3( x + 2)( x − 4).

We notice that k′ vanishes only when x = −2 or x = 4. These are the only candidates for local maxima or minima. The second derivative is k″ ( x ) = 6 x − 6. Now k″ (4) = 18 > 0, so x = 4 is the location of a local minimum. Also k″ (−2) = −18 < 0, so x = −2 is the location of a local maximum. A glance at the graph of this function, as depicted in Fig. 3.10 , confirms our calculations.

Example 2

Find all local maxima and minima of the function g ( x ) = x + sin x .

Solution 2

First we calculate that

g′ ( x ) = 1 + cos x .

Thus g′ vanishes at the points (2 k + 1) for k = ..., −2, −1, 0, 1, 2, .... Now g″ ( x ) = sin x . And g″ ((2 k + 1) ) = 0. Thus the second derivative test is inconclusive. Let us instead look at the first derivative. We notice that it is always ≥ 0. But, as we have already noticed, the first derivative changes sign at a local maximum or minimum. We conclude that none of the points (2 k + 1) is either a maximum nor a minimum. The graph in Fig. 3.11 confirms this calculation.

Example 3

A box is to be made from a sheet of cardboard that measures 12″ × 12″. The construction will be achieved by cutting a square from each corner of the sheet and then folding up the sides (see Fig. 3.12 ). What is the box of greatest volume that can be constructed in this fashion?

Fig. 3.12

Solution 3

It is important in a problem of this kind to introduce a variable. Let x be the side length of the squares that are to be cut from the sheet of cardboard. Then the side length of the resulting box will be 12 − 2 x (see Fig. 3.13 ). Also the height of the box will be x . As a result, the volume of the box will be

V ( x ) = x · (12 − 2 x ) · (12 − 2 x ) = 144 x − 48 x ² + 4 x ³ .

Our job is to maximize this function V .

Fig. 3.13

Now V′ ( x ) = 144 − 96 x + 12 x ² . We may solve the quadratic equation

144 − 96 x + 12 x ² = 0

to find the critical points for this problem. Using the quadratic formula, we find that x = 2 and x = 6 are the critical points. Now V″ ( x ) = −96 + 24 x . Since V″ (2) = −48 < 0, we conclude that x = 2 is a local maximum for the problem. In fact we can sketch a graph of V ( x ) using ideas from calculus and see that x = 2 is a global maximum.

We conclude that if squares of side 2″ are cut from the sheet of cardboard then a box of maximum volume will result.

Observe in passing that if squares of side 6″ are cut from the sheet then (there will be no cardboard left!) the resulting box will have zero volume. This value for x gives a minimum for the problem.

Example 4

A rectangular garden is to be constructed against the side of a garage. The gardener has 100 feet of fencing, and will construct a three-sided fence; the side of the garage will form the fourth side. What dimensions will give the garden of greatest area?

Solution 4

Look at Fig. 3.14 . Let x denote the side of the garden that is perpendicular to the side of the garage. Then the resulting garden has width x feet and length 100 − 2 x feet. The area of the garden is

A ( x ) = x · (100 − 2 x ) = 100 x − 2 x ² .

Fig. 3.14

We calculate A′ ( x ) = 100 − 4 x and find that the only critical point for the problem is x = 25. Since A″ ( x ) = −4 for all x , we determine that x = 25 is a local maximum. By inspection, we see that the graph of A is a downward-opening parabola. So x = 25 must also be the global maximum that we seek. The optimal dimensions for the garden are

width = 25 ft. length = 50 ft.

The sum of two positive numbers is 60. How can we choose them so as to maximize their product?

Solution 5

Let x be one of the two numbers. Then the other is 60 − x . Their product is

P ( x ) = x · (60 − x ) = 60 x − x ² .

Thus P is the quantity that we wish to maximize. Calculating the derivative, we find that

P′ ( x ) = 60 − 2 x .

Thus the only critical point for the problem is x = 30. Since P″ ( x ) ≡ −2, we find that x = 30 is a local maximum. Since the graph of P is a downward-opening parabola, we can in fact be sure that x = 30 is a global maximum.

We conclude that the two numbers that add to 60 and maximize the product are 30 and 30.