Universiti Pendidikan Sultan Idris

1. Beginning Calculus
Applications of Di¤erentiation
- Maxima and Minima Problems -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,
FSMT - UPSI
(LECTURE SLIDES SERIES)
VillaRINO DoMath, FSMT-UPSI
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2. Maximum and Minimum Problems
Learning Outcomes
Finding the critical points, critical values and end points.
Determine the optimal solutions.
VillaRINO DoMath, FSMT-UPSI
(DA2) Applications of Di¤erentiation - Maxima and Minima Problems 2 / 12

3. Maximum and Minimum Problems
Finding Maximum and Minimum Graphically
Max
Min
Finding the maximum and minimum of a graph of a function is easy
when the graph is sketched, but sketching the graph is time
consuming.
VillaRINO DoMath, FSMT-UPSI
(DA2) Applications of Di¤erentiation - Maxima and Minima Problems 3 / 12

4. Maximum and Minimum Problems
Keys to Finding Max/Min
Critical points
End Points
Points of discontinuity.
VillaRINO DoMath, FSMT-UPSI
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5. Maximum and Minimum Problems
Example
Problem: A wire of length 1 is cut into two pieces. Each piece encloses
a square. Find the largest area enclosed.
Draw a diagram and name the variables.
1 unit length
x 1 - x
x/4 (1 –x)/4
Area:
A (x) =
x
4
2
+
1 x
4
2
VillaRINO DoMath, FSMT-UPSI
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6. Maximum and Minimum Problems
Example - continue
Find the critical points, set A0 = 0 :
A0
(x) =
x
8
1 x
8
= 0 , x =
1
2
The critical value:
A
1
2
=
1
64
+
1
64
=
1
32
The endpoints, between 0 < x < 1.
lim
x!0+
A (x) =
1
16
, lim
x!1
A (x) =
1
16
VillaRINO DoMath, FSMT-UPSI
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7. Maximum and Minimum Problems
Example - continue
Minimum area enclosed is A
1
2
=
1
32
(these are equal squares) -
the minimum value.
Maximum area enclosed is A (0) =
1
16
or A (1) =
1
16
- the
maximum values.
The minimum value occurs at x =
1
2
, and the maximum values
occurs at x near 0 or x near 1.
Alternatively, the minimum point is
1
2
,
1
32
, and the maximum
points are 0,
1
16
and 1,
1
16
.
VillaRINO DoMath, FSMT-UPSI
(DA2) Applications of Di¤erentiation - Maxima and Minima Problems 7 / 12

8. Maximum and Minimum Problems
Example
Problem: Find a box (with square bottom) without a top with least
surface area for a …xed volume.
Draw the diagram and name the variables:
x
x
y
Volume:
V = x2
y, y …xed, volume …xed
Surface Areas:
A = 4xy + x2
VillaRINO DoMath, FSMT-UPSI
(DA2) Applications of Di¤erentiation - Maxima and Minima Problems 8 / 12

9. Maximum and Minimum Problems
Example - continue
y =
V
x2
. Then,
A (x) = 4
V
x
+ x2
The critical point(s):
A0
(x) =
4V
x2
+ 2x = 0 , x = (2V )1/3
The end point(s), between 0 < x < ∞ :
lim
x!0+
A (x) = ∞, lim
x!∞
A (x) = ∞
Using the second derivative (to test the concavities)
A00
(x) =
8V
x3
+ 2 > 0 for 0 < x < ∞ concave up
VillaRINO DoMath, FSMT-UPSI
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10. Maximum and Minimum Problems
Example - continue
The least surface area occurs at x = 21/3V 1/3 and y = 2 2/3V 1/3
The surface area is
A (x) = 4
V
21/3V 1/3
+ 21/3
V 1/3
2
= 3 21/3
V 2/3
VillaRINO DoMath, FSMT-UPSI
(DA2) Applications of Di¤erentiation - Maxima and Minima Problems 10 / 12

11. Maximum and Minimum Problems
Example - continue - More meaningful answer
Dimesionless variable
A
V 2/3
= 3 21/3
The ratio:
x
y
=
21/3V 1/3
2 2/3V 1/3
= 2
the length x is twice the height. The optimal shape of the box.
VillaRINO DoMath, FSMT-UPSI
(DA2) Applications of Di¤erentiation - Maxima and Minima Problems 11 / 12

12. Maximum and Minimum Problems
Example - continue - Using implicit di¤erentiation
V = x2y and A = 4xy + x2
d
dx
(V ) =
d
dx
x2
y
0 = 2xy + x2
y0
) y0
=
2y
x
d
dx
(A) =
d
dx
4xy + x2
0 = 4y + 4xy0
+ 2x = 4y + 4x
2y
x
+ 2x = 2x 4y
)
x
y
= 2
Advantage: nicer and faster than previous method.
Disadvantage: did not check whether the critical point(s).
VillaRINO DoMath, FSMT-UPSI
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