4.7 Solving Max-Min Problems
1. Read3. Identify the known quantities and the
unknowns. Use a variable.
2. Identify the quantity to be optimized. Write a
model for this quantity. Use appropriate
formulas. This is the primary function.
3. If too many variables are in the primary function
write a secondary function and use it to
eliminate extra variables.
4. Find the derivative of the primary function.
5. Set it equal to zero and solve.
6. Reread the problem and make sure you have
answered the question.
Figure 3.43: An open box made by cutting the corners
An open box is to be made by cutting
from a square sheet of tin. (Example 1)
squares from the corner of a 12 by 12 inch
sheet and bending up the sides. How large
should the squares be cut to make the box
hold as much as possible?
Figure
open
box
cuttingfrom
the the
corners
An open3.43:
box isAn
to be
made
bymade
cuttingbysquares
corner of a
from
sheet
ofbending
tin. (Example
1) How large should the
12 by a
12square
inch sheet
and
up the sides.
squares be cut to make the box hold as much as possible?
Maximize the volume
V=lwh
V = (12 – 2x) (12 – 2x) x =144x – 48x2 + 4x3
V  = 144 – 96x + 12x2 = 12(12 –8x + x2)
12(12 –8x + x2) = 0
(6-x)(2-x) = 0
x = 6 or x = 2
V   = -92 + 24x is negative at x = 2. There is a
relative max. Box is 8 by 8 by 2 =128 in3.
Figure 3.46: The graph of A = 2 r + 2000/r is concave up.
2
Minimizing surface area
You have been asked to design a 1 liter oil can in
the shape of a right cylinder. What dimensions
will use the least material?
Figure
The tograph
= 2oil
 r(1 liter
+ 2000/r
is concave
up.
3) can in the
You
have3.46:
been asked
designofa 1Aliter
= 1000cm
shape of a right cylinder. What dimensions will use the least material?
2
Minimize surface area
1000
S  2 r  2 rh where 1000   r h  h 
2
r
1000
2
2
1
S  2 r  2 r
 2 r  2000r
2
r
S   4 r  2000r 2
2000
4 r  2  0
r
4 r 3  2000
Use the 2nd derivative test
500
r3
 5.42
to show values give local

minimums.
h  10.84
2
2
4.8 Business Terms
C
x = number of items
p = unit price
C = Total cost for x items
R = xp = revenue for x items
= average cost for x units
P = R – C or xp - C
The daily cost to manufacture x items is C = 5000 + 25x 2.
How many items should manufactured to minimize the
average daily cost.
5000
C
 25 x
x
C  

5000
5000
x
2
x
2
 25
 25  0
5000  25x  0
2
x  200  14.14
14 items will
minimize the daily
average cost.
4.10
Old problem
Given a function, find its derivative
function
derivative
Inverse problem
Given the derivative, find the function.
.
Find a function that has a derivative y = 3x2
The answer is called the antiderivative
You can check your answer by differentiation
Curves with a derivative of 3x2
Each of these curves is
an antiderivative of y = 3x2
Antiderivatives
Derivative
y  x n
y  kf ( x)
y  f ( x)  g ( x)
Antiderivative
x n 1
y
 C , n  1
n 1
y  kf ( x)  C
y  f ( x)  g ( x) 
Find an antiderivative
3x 4  5 x 2  x  2
x 1
x
 3x
2
2

2
Find antiderivatives
y  3x4  5x2  x  2
x5
x3 x 2
y 3 5 
 2x  C
5
3
2
3 5 5 3 1 2
y  x  x  x  2x  C
5
3
2
Check by differentiating
Find an antiderivative
1
x2
x 1 x 1
y 
 1  
x
2
x
3
1
2 2
y  x  2x 2  C
3

y  3 x  2
2

2
1

x 2
 9 x  12 x  4
4
x5
x3
y  9  12  4 x  C
5
3
2
9 5
y  x  4 x3  4 x  C
5
Trigonometric derivatives
d
 sin x   cos x
dx
d
 cos x    sin x
dx
d
 tan x   sec2 x
dx
d
 cot x    csc2 x
dx
d
 sec x   sec x tan x
dx
d
 csc x    csc x cot x
dx
Derivative
Antiderivative
y  cos x
y  sin x  C
y  sin x
y   cos x  C
y  sec2 x
y  tan x  C
y  csc2 x
y   cot x  C
y  sec x tan x
y  sec x  C
y  csc x cot x
y   csc x  C