MCV4U | U n i t 3 – C u r v e S k e t c h i n g
Name: _____________
Date: _____________
3.6 – Optimization Part 1
In many real-life problems and situations, it is necessary to find the absolute best or optimal (largest, smallest) value.
We will need to find absolute max/min over a given interval to do this!
STEPS:
1)
2)
3)
4)
5)
Define the variables and express each with respect to one variable
Write equations and state restrictions
Differentiate the quantity to be optimized
Set 𝑓′(𝑥) = 0 to find CPs
Check CPs and endpoints to determine the absolute max/min values
Example 1: A flu epidemic breaks out. The fraction of the population infected at time t, in weeks, is given by
𝑓(𝑡) = (
64𝑡
. What is the largest fraction of the population that is infected thin
after the first 12 weeks?
8+𝑡)3
Restrictions:
f(t) 64t
(8 t)3
=
t 12
=
0
+
=
+ 7
f(x) 64(8 +)3
64(8
t = 12
0
+
-
(8
=
8
3
3/8 t)2(64t)
+
=
-
t)6
+
t)" [(8 t)
-
+
+
3t]
(8 t)b
+
64(8 2t)or
-
-
=
f(x)
(8 t)4
0
-
4)
(+ 5)4
+
+
->(X)
128(t
=
->'(X) DNE
0
=
64(8
=
2t)(8 t)" 0
=
+
-
t
0 8 -2t
-
=
8
M
=
Reject
2t 8
=
(Noti n
t 4
domain)
=
Endpoints:
CPS:
-> (4)
64(4)
=
f(0)
6430)
=
f(12) 64(12)
=
(8 12)3
(8 0)3
(8 4)3
+
+
+
I
O
=
=
27
⑭
MIN
Absolute
0.148
=
-5
0.096
=
p
absolute
MAX
·
The
population
of
largestfraction
which
occurs
after
4
is
that
infected
is
at
weeks.
3 . 6 – O p t i m i z a t i o n P a r t 1 | 15
MCV4U | U n i t 3 – C u r v e S k e t c h i n g
Name: _____________
Date: _____________
Example 2: Find two positive integers such that their product is 192 and the sum of the first integer and three times the
second integer is a minimum.
1st
the
x
represent
Let
representthe
Lety
xy
y
the
represent
Let's
S
192
integer
2nd
integer.
of
the
sum
3y
x
2
x
=
THIS
⑭
we
3(2)
x
=
integer.
function
the
is
optimizing!
are
+
=
S(x)
and
the
times
+
=
=
S(x)
3
1st
and
Restrictions
76
+
1 x +
=
192,
Xt
S(X)
S'(x) 1
=
=
-
1
x
576x
x2
S'(x)
54
-
x
-
+
57642
-
=
=
x
1
S'(X) DNE
0
=
2
576
-
x
0
z
0
=
=
x 0
=
(X 24)(X 24)
+
-
x
2
x
24
-
=
↑
Reject
↓
↓
576
0
=
24
=
A
Reject
CP's:
S(24)
24
=
Endpoints:
S
+
S(1)
48
1
=
>(192)
56
+
192
=
=
577
=
AbSOlUH
·.
x
96
+
195
=
*
MIR
Absolute
Max
24
=
S(24)
y
48
=
=
·
The
the
8
y
8
two
integers
minimum
and
that
produce
sum
are
24.
=
3 . 6 – O p t i m i z a t i o n P a r t 1 | 16
MCV4U | U n i t 3 – C u r v e S k e t c h i n g
Name: _____________
Date: _____________
Example 3: A train carries 2000 passengers daily for a ticket price of $7. For every $0.10 increase in price, 40 fewer
passengers will travel (and vice versa). If the train has a capacity of 2600 people, determine the ticket price that will
maximize the revenue.
i
Let
representthe
R
Let
the
represent
)
R(n)
Price
=
number
Revenue
(7 0.1n)(2000
R(n)
tickets
of
40n)
=
+
R(n) 14000-280n
=
+200n
-
increases.
price
dollars.
in
ticket) (#
per
$0.10
of
-4n2
sold)
Restrictions:
①
#
passengers
of
2000
R(n)
=
40n = 2600
-
14000
4n2 -80n
-
+
-
40n 600
=
n 1
R'(n)
=
R'(n)
8n
-
-
80
②
#
2000
8(n 10)
-
-
+
-
15
=
40n 0
4On - 2000
n
↓
n
=
-
0
=
passengers
0
=
-
of
=
0
= 2600
10
:
-
50
=
151n 50
=
CP's:
R1
-
10)
[x
=
0.17
+
-
Endpoints:
10)3[2000
-
40(10))
R(
6(2400)
-
15) 14300
=
=
=14400
Ticketprice
7
=
A
max
price
0
0.1(-10)
+
revenue
ticketprice
(10
Revenue
=
6
=
·
Max
&
R(50)
$14400
of
occurs
at
a
$6.
of
decreases
of
$0.10).
3 . 6 – O p t i m i z a t i o n P a r t 1 | 17