To optimize something means to maximize or minimize
some aspect of it…
Strategy for Solving Max-Min Problems
1. Understand the Problem. Read the problem carefully.
Identify the information you need to solve the problem.
2. Develop a Mathematical Model of the Problem. Draw
pictures and label the parts that are important to the
problem. Introduce a variable to represent the quantity
to be maximized minimized. Using that variable, write a
function whose extreme value gives the information
sought.
3. Graph the Function. Find the domain of the function.
Determine what values of the variable make sense in the
problem.
To optimize something means to maximize or minimize
some aspect of it…
Strategy for Solving Max-Min Problems
4. Identify the Critical Points and Endpoints. Find where
the derivative is zero or fails to exist.
5. Solve the Mathematical Model. If unsure of the result,
support or confirm your solution with another method.
6. Interpret the Solution. Translate your mathematical
result into the problem setting and decide whether the
result makes sense.
Guided Practice
An open-top box is to be made by cutting congruent squares of
side length x from the corners of a 20- by 25-inch sheet of tin and
bending up the sides. How large should the squares be to make
the box hold as much as possible? What is the resulting max.
volume?
First, sketch a diagram…
Dimensions of the box?
Volume of the box?
x
by
20  2x
by
25  2x
V  x   x  20  2 x  25  2 x 
 4 x  90 x  500 x
3
Domain restriction?
D :  0,10
2
Guided Practice
An open-top box is to be made by cutting congruent squares of
side length x from the corners of a 20- by 25-inch sheet of tin and
bending up the sides. How large should the squares be to make
the box hold as much as possible? What is the resulting max.
volume?
Graphical solution?
Analytic solution?
 Graph V
 x , calc. the maximum
2

V  x   12x 180 x  500
CP: x  3.681,11.319
Check Endpoints and Critical Points:
V  0  0 V 10  0 V  3.681  820.528
Guided Practice
An open-top box is to be made by cutting congruent squares of
side length x from the corners of a 20- by 25-inch sheet of tin and
bending up the sides. How large should the squares be to make
the box hold as much as possible? What is the resulting max.
volume?
Interpretation?
Cutting out squares that are roughly 3.681 inches
on a side yields a maximum volume of
approximately 820.528 cubic inches.
Guided Practice
You have been asked to design a one-liter oil can shaped like
a right circular cylinder. What dimensions will use the least
material?
Sketch a diagram…
Volume (in cm 3 ):
1000   r h
Surface Area (in cm 2 ):
2
A  2 r  2 rh
2
We need to find a radius and height that will minimize the
surface area, all while satisfying the constraint given by
the volume…
Guided Practice
You have been asked to design a one-liter oil can shaped like
a right circular cylinder. What dimensions will use the least
material?
1000   r h
2
A  2 r  2 rh
2
1000
h
2
r
2000
 1000 
2
A  2 r  2 r 
 2 r 
2 
r
 r 
2
Guided Practice
You have been asked to design a one-liter oil can shaped like
a right circular cylinder. What dimensions will use the least
material?
A is differentiable, with r > 0, on an interval with no endpoints.
Check the first derivative:
2000
2000
A  2 r 
A  4 r  2
r
r
2000
0  4 r  2
r
500
3
So, what happens
r3
4 r  2000
2

at this r-value?
Guided Practice
You have been asked to design a one-liter oil can shaped like
a right circular cylinder. What dimensions will use the least
material?
2000
A  2 r 
r
2
2000
A  4 r  2
r
Check the second derivative:
4000
A  4  3
r
This function is positive for all
positive r-values, so the graph
of A is concave up for the entire
domain…
Which means our r-value
signifies an abs. min.
r
3
500

Guided Practice
You have been asked to design a one-liter oil can shaped like
a right circular cylinder. What dimensions will use the least
material?
Solve for height:
1000
h
2
r
r
3
500

Interpretation:
The one-liter can that uses the least material has
height equal to the diameter, with a radius of
approximately 5.419 cm and a height of
approximately 10.839 cm.
Guided Practice
A rectangle has its base on the x-axis and its upper two vertices
2
on the parabola y  12  x . What is the largest area the
rectangle can have, and what are its dimensions?
Model the situation:
Dimensions of the rectangle:
 x,12  x 
2
2x
by
12  x
2
Area of the rectangle:
A  2 x 12  x
 24 x  2 x
Domain:

2

3
D : 0, 12

Guided Practice
A rectangle has its base on the x-axis and its upper two vertices
2
on the parabola y  12  x . What is the largest area the
rectangle can have, and what are its dimensions?
Maximize analytically:
Model the situation:
 x,12  x 
2
A  x   24  6x
2
 64  x
CP:
2

x2
This corresponds to a
maximum area (…Why?)
Largest possible area is A(2) = 32,
dimensions are 4 by 8.
Guided Practice
A 216-m2 rectangular pea patch is to be enclosed by a fence
and divided into two equal parts by another fence parallel to
one of the sides. What dimensions for the outer rectangle will
require the smallest total length of fence? How much fence
will be needed?
Area of the patch:
Model the situation:
216
x
A  xy  216  y 
Total length of the fence:
y
x
x
x
y L  x   3x  2 y  3x  2  216 


x 

1
 3 x  432 x
D :  0,  
Guided Practice
A 216-m2 rectangular pea patch is to be enclosed by a fence
and divided into two equal parts by another fence parallel to
one of the sides. What dimensions for the outer rectangle will
require the smallest total length of fence? How much fence
will be needed?
Maximize analytically:
2

L  x   3  432 x
432
3 2  0
x
2
3 x  432
CP:
x  12
This corresponds to a
minimum length (…Why?)
The pea patch will
measure 12 m by 18 m
(with a 12-m divider), and
the total amount of fence
needed is 72 m.