3.6 applications in optimization

Applications in Optimization
In this section, we solve various extrema problems
in real world applications. These problems are
referred to as “optimization problems” in
mathematics, and derivatives play a major role in
such problems.

such problems. We begin with an
extrema-existence–problem.

Let y be a function defined over an
interval V, y may or may not have
any extremum in V.

any extremum in V. Here are two
examples of well behaved functions
that don’t have extrema.

Let y = tan(x) over the interval
V = (–π/2, π/2) as shown.
y=tan(x)
x
y
–π/2 π/2

Let y = tan(x) over the interval
V = (–π/2, π/2) as shown. There is no
extremum in V because V is open.
y=tan(x)
x
y
–π/2 π/2

Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.

x
y=g(x)
For example, let y = g(x) be the
function over the closed interval
U = [–π/2, π/2] as shown.
y
(0, 1)
–π/2 π/2
(0, –1)

x
y=g(x)
U = [–π/2, π/2] as shown. Note that
even though g(x) is bounded between
y
(0, 1)
–π/2 π/2
(0, –1)
1 and –1, it still doesn’t have any extremum in U.

x
y=g(x)
y
(0, 1)
–π/2 π/2
(0, –1)
(Extrema Theorem for Continuous Functions)

x
y=g(x)
y
(0, 1)
–π/2 π/2
(0, –1)
Let y = f(x) be a continuous function defined over a
closed interval V = [a, b], then both the absolute max.
and the absolute min. exist in V.

x
y=g(x)
y
(0, 1)
–π/2 π/2
(0, –1)
and the absolute min. exist in V. Furthermore, the
absolute extrema must occur where f '(x) = 0,

x
y=g(x)
y
(0, 1)
–π/2 π/2
(0, –1)
and the absolute min. exist in V. Furthermore, the
absolute extrema must occur where f '(x) = 0, or where
f'(x) is UDF, or they occur at the end points {a, b}.

Here are examples of each type of extrema.

A.
a
x
c b

A.
a
x
b
In A. The absolute maximum
occurs at x = c where
f '(c) = 0, and the absolute
minimum occurs at the end
point x = b.
c
A f'=0 ab. max.
An end–point
ab. min.

A.
a
An end–point
ab. min.
x
A f'=0 ab. max.
b
B.
a b x
point x = b.
c
c d

A.
a
x
b
B.
a b x
point x = b.
c
c d
In B. The absolute maximum
f '(c) is UDF, and the absolute
minimum occurs at x = d
where f '(d) = 0.
A f'=0 ab. max.
An end–point
ab. min.
A UDF– f' ab. max.
A f'=0
ab. min.

A.
a
x
b
B.
a b x
point x = b.
c
c d
In B. The absolute maximum
f '(c) is UDF, and the absolute
minimum occurs at x = d
where f '(d) = 0.
A f'=0 ab. max.
An end–point
ab. min.
A UDF– f' ab. max.
A f'=0
ab. min.
Hence we have to consider
points of all three cases when
solving for extrema.

Example A. Let y be defined as the following.
x2 for 1 > x
y =
2 – x for 1 ≤ x

x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y

x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
Find the extrema of y over the
interval [–1 , ½ ].

x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
The end–point values are
f(–1) = 1 and f(½) = ¼.

x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
x
y
f(–1) = 1 and f(½) = ¼.
(–1, 1)
–1 ½
(½, ¼ )

x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
x
y
f(–1) = 1 and f(½) = ¼.
y' = 0 at x = 0 so that (0, 0)
is an f'=0–type minimum.
(–1, 1)
–1 ½
(½, ¼ )

x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
x
y
f(–1) = 1 and f(½) = ¼.
y' = 0 at x = 0 so that (0, 0)
is an f'=0–type minimum.
Within the interval [–1 , ½ ], there
is no point of the f'–UDF–type.
Hence the ab. max. is at (–1, 1)
and the ab. min is at (0, 0).
(–1, 1)
–1 ½
(½, ¼ )

The key here is that if we are to find extrema of a
given continuous function over a closed interval
we have to obtain all three types of points described
above, then select the correct solutions.

x
For instance, within [–½, 3], y
x = 0 is an f'=0–type solution.
The point (1, 1) at x = 1 is an
f'–UDF–type solution.
(1, 1)
–½ 3

x
f(–½) = ¼ and f(3) = –1.
(1, 1)
–½ 3
(3, –1)

x
f(–½) = ¼ and f(3) = –1.
By inspection, the ab. max. is at
(1, 1) and the ab. min is at (3, –1).
(1, 1)
–½ 3
(3, –1)

x
f(–½) = ¼ and f(3) = –1.
By inspection, the ab. max. is at
(1, 1) and the ab. min is at (3, –1).
(1, 1)
–½ 3
(3, –1)
In the applied problems below, make sure all such
relevant points are considered.

Distance Problems

Distance Problems
The distance formula is a square–root formula whose
radicand is always nonnegative. To find the extrema
of a distance relation in the form of “√u(x)”,
we drop the square–root and look for the extrema of
u(x) instead.

Distance Problems
u(x) instead.
Example B. Find the coordinate of the point on the
line y = 2x that is closest to the point (1, 0).

Distance Problems
u(x) instead.
x
y=2x
(1, 0)
y
D

Distance Problems
u(x) instead.
x
y=2x
(1, 0)
y
(x, 2x)
A generic point on the line is (x, 2x)
and its distance to the point (1, 0)
is D = [(x – 1)2 + (2x – 0 )2]½.
D

Distance Problems
u(x) instead.
x
y=2x
(1, 0)
y
(x, 2x)
A generic point on the line is (x, 2x)
and its distance to the point (1, 0)
is D = [(x – 1)2 + (2x – 0 )2]½.
We want to find the x value that
gives the minimal D. Note that D is
defined for all numbers.
D

x
(1, 0)
(x, 2x)
We minimize the radicand
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
y
D

x
(1, 0)
(x, 2x)
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
and the point is (1/5, 2/5).
y
D

x
(1, 0)
(x, 2x)
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
From the geometry it’s obvious
(1/5, 2/5) is the closest point to (1, 0).
y
D

x
(1, 0)
(x, 2x)
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
½
y
D
Similarly if we wish to find the extrema of eu(x) or
In(u(x)), we may find the extrema of u(x) instead.
This is true because like sqrt(x), eu(x) and In(u(x)) are
increasing functions, i.e. if s < t, then es < et,
so the extrema of eu(x) are the same as u(x).

x
(1, 0)
(x, 2x)
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
½
y
D
Similarly if we wish to find the extrema of eu(x) or
In(u(x)), we may find the extrema of u(x) instead.
This is true because like sqrt(x), eu(x) and In(u(x)) are
increasing functions, i.e. if s < t, then es < et,
so the extrema of eu(x) are the same as u(x).
In fact, if f(x) is an increasing function, then the
extrema of f(u(x)) are the same as the ones of u(x).

(What is the corresponding statement about the
extrema of f(u(x)) and u(x) if f(x) is a decreasing
function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)

Area Problems
In area optimization problems, the main task is to
express the area in question in a single well chosen
variable x as A(x).

Area Problems
variable x as A(x).
If x represents a specific measurement, as opposed to
a coordinate, then x must be nonnegative (often it’s
also bounded above).

Area Problems
variable x as A(x).
If x represents a specific measurement, as opposed to
a coordinate, then x must be nonnegative (often it’s
also bounded above). In such cases consider the
significance when x = 0 (or it’s other boundary value).
Often one of the extrema is at the boundary, and we
are looking for the other extremum.
A drawing is indispensable in any geometric problem.

Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
to build rectangular regions as
shown to separate the animals.
What is the maximal enclosed
area that is possible? x
x/2
y

area that is possible?
The enclosed area is A = xy.
x
x/2
y

x/2
y
x
The enclosed area is A = xy. The total linear length
is 120, that is, 3y + 2½ x = 120

x/2
y
x
is 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6.

x/2
y
x
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.

x/2
y
x
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.
At the boundary values x = 0, 48 (why?) we have the
absolute minimal A = 0.

x/2
y
x
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.
absolute minimal A = 0. Setting A'(x) = 40 – 5x/3 = 0,
we get x = 24

x/2
y
x
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.
absolute minimal A = 0. Setting A'(x) = 40 – 5x/3 = 0,
we get x = 24 which yields y = 20.
So 20×24 = 480 yd2 must be the maximum area.

In fact for any specified partition of the rectangle, the
maximal area always occurs when the total–vertical–
fence–length is equal to the total–horizontal–fence–
length. That is, using half of the fence for vertical
dividers and the other half for the horizontal dividers
will give the maximum enclosed area.
Hence if farmer Joe is to
build an enclosure with 120
yd of fence as shown,
x/2
then the answer is
y
x = 60/3½ = 120/7,
y = 60/3 = 20
with the maximum possible
area of 120/7×20 ≈ 343 yd2. x

Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
120 yd3
h
r

120 yd3
h
r
The volume of a cylinder is V = Bh
where B = area of the base
and h = height

120 yd3
h
r
and h = height so we have120π = πr2h
or h = 120/r2.

120 yd3
h
r
or h = 120/r2. The surface area of the
cylinder consists of a circular
base and the circular wall.

120 yd3
base and the circular wall. The base area is πr2.
h
r

120 yd3
h
r
Unroll and flatten the wall, it is an h x 2πr rectangular
sheet.

120 yd3
h
r
sheet. Hence the total surface is S = πr2+ 2πrh,
or S = πr2 + 2πr( 1 2 0 )
r2

120 yd3
h
r
sheet. Hence the total surface is S = πr2+ 2πrh,
or S = πr2 + 2πr( 1 2 0 ) = πr2 + 240π/r = π(r2 + 240/r)
r2

120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
dividing both sides by 2π and clearing
the denominator yields r3 – 120 = 0

120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
or that r = 1201/3, which gives the
absolute minimal (why?) surface

120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
of π(1202/3 + ) = π 360/120 240 1/3
1201/3 yd2.

120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
of π(1202/3 + ) = π 360/120 240 1/3
1201/3 yd2.
If the geometric object has a variable angle θ, then θ
may be utilized as the independent variable.

120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
of π(1202/3 + ) = π 360/120 240 1/3
1201/3 yd2.
If the geometric object has a variable angle θ, then θ
may be utilized as the independent variable.
Example E.
Find the isosceles triangle
4
having two sides of length
4
4 with the largest area as
shown.

There are two observations that
will make the algebra cleaner. 4
4

will make the algebra cleaner.
First let’s take advantage of the
symmetry. Draw a perpendicular
line as shown, it cuts the triangle
4
4
into two mirror image right
triangles.

4
4
triangles. We may maximize one
of the right triangles instead,
i.e. to find the largest area
possible of a right triangle with
hypotenuse equal to 4.
4
a
b A

4
4
triangles. We may maximize one
of the right triangles instead,
i.e. to find the largest area
possible of a right triangle with
hypotenuse equal to 4.
4
a
b
A
θ
Second, instead of expressing the area A in terms of
the measurements of the legs a and b,
we express A in terms of the angle θ as shown.

A = ½ ab
so in terms of θ we have
4
a
θ
b A
A = ½ 4sin(θ) 4cos(θ)
= 8 sin(θ) cos(θ)

A = ½ ab
Note that 0 ≤ θ ≤ π/2.
4
a
θ
b A
= 8 sin(θ) cos(θ)

A = ½ ab
Set A'(θ) = 8(cos2(θ) – sin2 (θ)) = 0
4
a
θ
b A
= 8 sin(θ) cos(θ)

A = ½ ab
Set A'(θ) = 8(cos2(θ) – sin2 (θ)) = 0
4
a
θ
b A
= 8 sin(θ) cos(θ)
Hence cos(θ) = ±sin (θ)
However, the only solution with 0 ≤ θ ≤ π/2 is θ = π/4

A = ½ ab
Set A'(θ) = 8(cos2(θ) – sin2 (θ)) = 0
4
a
b
Hence cos(θ) = ±sin (θ)
which must be a maximum (why?).
So the largest possible isosceles
triangle in question is the right
isosceles triangle as shown with
area 2A = ½ (4 ×4) = 8.
A
θ
= 8 sin(θ) cos(θ)
However, the only solution with 0 ≤ θ ≤ π/2 is θ = π/4
4
4 π/2

Example F. A duck can walk 3 ft/sec and swim
2 ft/sec at its top speed. It sees a piece of bread
across the pool as shown. What is the path it should
take to get to the bread as soon as possible?
40 ft
pool
30 ft

40 ft
pool Select the x as shown in the
picture.
30 ft
x

40 ft
picture. (The selection of the
correct x is important. The wrong
choice leads to tangled algebra
as is often the case in these
problems.)
30 ft
x

40 ft
problems.)
30 ft
Hence if your choice of x
leads to nowhere,
try another choice.
x

40 ft
pool
30 ft
x
Select the x as shown in the
problems.)
Note that 0 ≤ x ≤ 40, where x = 0
means the duck walks around
the pool and x = 40 means the duck swims all the way.

40 ft
pool
30 ft
x
(x2 + 900)½
For x ≠ 0, the duck swims a
distance of (x2 + 900)½

40 ft
pool
30 ft
x
(x2 + 900)½
40 – x
distance of (x2 + 900)½ and
walks a distance of (40 – x),

40 ft
pool
30 ft
x
and the total time t it takes is
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
swimming time walking time

40 ft
pool
30 ft
x
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0

40 ft
pool
30 ft
x
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½

40 ft
pool
30 ft
x
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½
3x = 2(x2 + 900)½

40 ft
pool
30 ft
x
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½
3x = 2(x2 + 900)½
9x2 = 4(x2 + 900)

40 ft
pool
30 ft
x
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½
3x = 2(x2 + 900)½
9x2 = 4(x2 + 900)
5x2 = 3600  x = ±12√5
Discarding the negative answer, x =12√5 ≈ 26.8 ft.

When x = 0, the time it takes is not t(0) because t(x)
assume the duck swims at least 30 ft,

assume the duck swims at least 30 ft, as it would be
the case if the duck is across a river from the bread
instead of a pool.

instead of a pool. Let’s examine the three points.

When x = 0, the duck walks 70 ft and it takes
70/3 = 23 1/3 seconds.
When x = 40 the duck swims all the way and it
takes t(40) = 50/2 = 25 seconds.
At the critical point, t(12√5) ≈ 24.5 seconds.

70/3 = 23 1/3 seconds.
Therefore the duck should walk around the pool.

70/3 = 23 1/3 seconds.
Therefore the duck should walk around the pool.
Remarks
1. As noted that if the duck is across a river instead
of a pool, then indeed x = 12√5 is the absolute
minimum because t(0) = swim + walk = 28 1/3 sec.

2. Just because the duck walks faster does not
automatically means it should always walk around
the pool. We may think about this by imagining the
duck is slightly faster in walking, say at 2.01 ft/sec,
than swimming at 2 ft/sec. Then of course the duck
should swim toward the bread instead of wasting
time walking around. In fact the closer the walking
speed is to the swimming speed, the more the duck
should swim toward the bread.
Back to math–265 pg

3.6 applications in optimization

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