Solving Optimization Problems

Absolute extrema

An optimization problem asks for the largest or smallest value of a quantity over every allowed input: an absolute extremum. A relative extremum isn't enough. In the graph below,

f

has a relative maximum at

x = 2

, and yet

f(7)

is larger.

On [0, 7], f has a relative maximum at x = 2, but its absolute maximum is f(7) = 7.7, at an endpoint. Its absolute minimum, f(0) = 0, is at the other endpoint.

Critical points, where the derivative is

0

or undefined, are the candidates. One of two arguments then shows which candidate gives the absolute extremum.

The lesson on the Second Derivative Test proves this. A second point where

f

beats

f(c)

would force a second critical point between them.

A closed domain

Example 1

A box with a square base is framed with steel tubing along all twelve edges. Tubing for the four edges of the base costs

3

dollars per foot, and tubing for the other eight edges costs

1

dollar per foot. The frame uses the whole budget of

48

dollars.

(a) Let the base be

x

feet on a side. Write the volume

V

as a function of

x

, and give its domain.

(b) Find the dimensions of the box with the largest volume, and that volume.

The frame. The four base edges (blue) use the tubing that costs 3 dollars per foot.

The volume V(x) = 12x² − 4x³ on [0, 3], largest at x = 2.

(a) Let

h

be the height in feet. The base edges cost

3 \cdot 4x = 12x

dollars. The top edges cost

4x

dollars and the vertical edges

4h

. The total is

16x + 4h = 48

, so

h = 12 - 4x

and

\begin{aligned} V(x) &= x^2 h = x^2(12 - 4x) \\[4pt] &= 12x^2 - 4x^3. \end{aligned}

The side

x

and the height

h = 12 - 4x

can't be negative, so

0 \le x \le 3

. At

x = 0

and

x = 3

the box is flat, with volume

0

. Including those two endpoints makes the domain the closed interval

[0, 3]

(b)

V

is a polynomial, so it's continuous on

[0, 3]

and differentiable on

(0, 3)

V'(x) = 24x - 12x^2 = 12x(2 - x).

(0, 3)

V'(x) = 0

only at

x = 2

. By the Candidates Test, compare

V

there and at the endpoints:

\begin{aligned} V(0) &= 0, \\[4pt] V(2) &= 2^2(12 - 8) = 16, \\[4pt] V(3) &= 3^2(12 - 12) = 0. \end{aligned}

The absolute maximum of

V

[0, 3]

16

, at

x = 2

, where

h = 12 - 4(2) = 4

. The box is

2

feet by

2

feet at the base and

4

feet tall, with a volume of

16

cubic feet.

An optimum at an endpoint

A limit on the variable can shut the critical point out of the domain. When no critical point lies inside a closed domain, the Candidates Test compares the endpoints alone, and the optimum is at one of them.

Example 2

The box in Example 1 has to fit under a shelf, so its height can be at most

3

feet. With the same tubing and the same

48

-dollar budget, find the dimensions of the box with the largest volume, and that volume.

The height is still

h = 12 - 4x

, and

12 - 4x \le 3

gives

x \ge \frac{9}{4}

. The domain is now

\left[\frac{9}{4}, 3\right]

. The critical point

x = 2

from Example 1 would need a height of

4

feet, so it's out.

For

\frac{9}{4} < x < 3

, the factor

12x

is positive and

2 - x

is negative, so

V'(x) = 12x(2 - x) < 0

. There's no critical point in the interval, and

V

is continuous on it. By the Candidates Test, compare the endpoints:

\begin{aligned} V\!\left(\tfrac{9}{4}\right) &= \left(\tfrac{9}{4}\right)^2(12 - 9) = \tfrac{243}{16}, \\[4pt] V(3) &= 0. \end{aligned}

With h ≤ 3, only the blue part of V is allowed. V decreases there, so its maximum is at the left endpoint x = 9/4.

The maximum is at

x = \frac{9}{4}

, where

h = 12 - 9 = 3

. The box is

2.25

feet by

2.25

feet at the base and

3

feet tall, as tall as the shelf allows. Its volume is

\frac{243}{16} \approx 15.19

cubic feet.

An unbounded domain

A domain such as

t \ge 0

has no right endpoint, so the Candidates Test doesn't apply. The one-critical-point argument still works there, with the First Derivative Test or the Second.

Example 3

To measure a stream's flow, a team releases dye upstream of a gauge. The concentration of dye at the gauge

t

minutes after the release is

C(t) = \frac{30t}{t^2 + 4t + 16}

milligrams per liter, for

t \ge 0

. Find the greatest concentration at the gauge and the time it occurs, with a justification.

The denominator

t^2 + 4t + 16 = (t + 2)^2 + 12

is always positive, so

C

is continuous and differentiable for

t \ge 0

. By the quotient rule, the numerator of

C'(t)

30\left(t^2 + 4t + 16\right) - 30t(2t + 4)

, which simplifies to

480 - 30t^2

. So

C'(t) = \frac{30\left(16 - t^2\right)}{\left(t^2 + 4t + 16\right)^2}.

C'(t)

is never undefined, and it's

0

only at

t = 4

in the domain (

t = -4

is outside it). For

0 \le t < 4

16 - t^2 > 0

, so

C'(t) > 0

. For

t > 4

16 - t^2 < 0

, so

C'(t) < 0

C'

changes from positive to negative at

t = 4

, so

C

has a relative maximum there by the First Derivative Test. It's the only critical point for

t > 0

, so

C

has its absolute maximum at

t = 4

C(4) = \frac{30(4)}{16 + 16 + 16} = \frac{120}{48} = 2.5.

The concentration C(t) rises to 2.5 mg/L at t = 4 minutes, then falls toward 0.

The concentration at the gauge is greatest

4

minutes after the release, at

2.5

milligrams per liter.

Example 4

Find the point on the curve

y = x^{3/2}

closest to the point

P(10, 6)

, and find that distance.

A point on the curve is

\left(x, x^{3/2}\right)

with

x \ge 0

. Its distance from

P

is the square root of

S(x)

below. The square root is increasing, so the distance is smallest where

S

is smallest. For

x \ge 0

, minimize

S(x) = (x - 10)^2 + \left(x^{3/2} - 6\right)^2.

S

is continuous on

[0, \infty)

and differentiable for

x > 0

, with

\begin{aligned} S'(x) &= 2(x - 10) \\ &\quad + 3x^{1/2}\left(x^{3/2} - 6\right) \\[4pt] &= 3x^2 + 2x - 18x^{1/2} - 20. \end{aligned}

Write

u = \sqrt{x}

, so

x = u^2

with

u \ge 0

. Then

\begin{aligned} S'(x) &= 3u^4 + 2u^2 - 18u - 20 \\[4pt] &= (u - 2)\,q(u), \end{aligned}

where

q(u) = 3u^3 + 6u^2 + 14u + 10

. For

u \ge 0

q(u) \ge 10 > 0

, so

S'(x)

has the sign of

u - 2

. Thus

S'(x) < 0

for

0 < x < 4

and

S'(x) > 0

for

x > 4

. By the First Derivative Test,

S

has a relative minimum at

x = 4

, its only critical point on

[0, \infty)

, so that's the absolute minimum:

S(4) = 36 + (8 - 6)^2 = 40

x = 4

y = 4^{3/2} = 8

. The closest point is

(4, 8)

, at a distance of

\sqrt{40} = 2\sqrt{10} \approx 6.325

from

P

The closest point Q(4, 8) on y = x^(3/2) to P(10, 6). The segment PQ meets the curve at a right angle.