This document discusses using calculus to find maximum and minimum values. It provides examples of finding critical points and values to determine optimal solutions for area and surface area problems. For the area problem, the minimum occurs at half the length with equal squares. For the surface area of a box problem, the least surface area occurs when the length is twice the height. The document demonstrates using differentiation and implicit differentiation to solve these types of applied optimization problems.

1. Beginning Calculus
Applications of Di¤erentiation
- Maxima and Minima Problems -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,
FSMT - UPSI
(LECTURE SLIDES SERIES)
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2. Maximum and Minimum Problems
Learning Outcomes
Finding the critical points, critical values and end points.
Determine the optimal solutions.
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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.
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4. Maximum and Minimum Problems
Keys to Finding Max/Min
Critical points
End Points
Points of discontinuity.
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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
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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
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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
.
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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
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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
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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
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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.
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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).
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