1. { ln x for 0 < x < 2 x2 ln 2 for 2 < x < 4 If f(x) =
Problem of the Day No calculator! {
ln x for 0 < x < 2
x2 ln 2 for 2 < x < 4
If f(x) =
then lim f(x) is
x⇒2
A) ln C) ln E) nonexistent
B) ln D) 4

2. limit from left does not equal limit from right at 2
Problem of the Day {
ln x for 0 < x < 2
x2 ln 2 for 2 < x < 4
If f(x) =
then lim f(x) is
x⇒2
A) ln C) ln E) nonexistent
B) ln D) 4
limit from left does not equal limit from right at 2

3. Optimization Problems
One of the most common uses of calculus is to determine minimum and maximum values. Some examples are
greatest profit
least cost
least time
greatest voltage
optimum size
greatest strength
greatest distance
greatest volume
least material
least size

4. Method 1 - Guess and check
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check h x x
surface area = area of base + 4 (area of a side)
108 = (x)(x) + 4(x)(h)
108 = x2 + 4xh

5. Method 1 - Guess and check
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check h x
108 = x2 + 4xh x
volume = x2h

6. Method 1 - Guess and check
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check h
108 = x2 + 4xh x x
volume = x2h
length of side
height
volume 3 4 5 6 7 8

7. Method 1 - Guess and check
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 1 - Guess and check h
108 = x2 + 4xh x x
volume = x2h
length of side height
volume 3 4 5 6 7 8
8.25
5.75
4.15 3
2.11
1.375
74.25 92
103 ¾
108
103¼ 88

8. Any problems with this method?
Method 1 - Guess and check h
108 = x2 + 4xh x x
volume = x2h
length of side height
volume 3 4 5 6 7 8
8.25
5.75
4.15 3
2.11
1.375
74.25 92
103 ¾
108
103¼ 88

9. What is this in Calculus?
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus h
surface area = area of base + 4 (area of a side) x x
108 = x2 + 4xh
volume = x2h
What is to be maximized?
What is this in Calculus?

10. What is this in Calculus? Critical Number
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
surface area = area of base + 4 (area of a side) h
108 = x2 + 4xh x x
volume = x2h
What is to be maximized? Volume
What is this in Calculus? Critical Number
And Critical Numbers come from?

11. What is this in Calculus? Critical Number
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
surface area = area of base + 4 (area of a side) h
108 = x2 + 4xh x x
volume = x2h
What is to be maximized? Volume
What is this in Calculus? Critical Number
And Critical Numbers come from?
First Derivative Set Equal to Zero

12. Method 2 - Use Calculus h How can we eliminate a variable?
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
surface area = area of base + 4 (area of a side) h
108 = x2 + 4xh x x
volume = x2h
How can we eliminate a variable?

13. Method 2 - Use Calculus h How can we eliminate a variable?
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
Method 2 - Use Calculus
surface area = area of base + 4 (area of a side) h
108 = x2 + 4xh x x
volume = x2h
How can we eliminate a variable?
Use the equation not being maximized to solve for one of the variables and substitute the result into the equation being maximized.

14. 2) Use equation not being maximized to eliminate a variable
Procedure to Maximizing or Minimizing
1) Develop 2 equations
2) Use equation not being maximized to
eliminate a variable
3) Determine feasible domain for x
4) Differentiate
5) Find critical numbers
6) Evaluate at endpoints of domain and
at critical numbers
7) Pick value that gives the largest or
smallest value

15. 2) Use equation not being maximized to eliminate a variable
1) Develop 2 equations
108 = x2 + 4xh
volume = x2h
2) Use equation not being maximized to
eliminate a variable
108 = x2 + 4xh
h = x2 4x
V = x2 h
V= (x2 ) x2 4x
V = 27x - ¼x3

16. 3) Determine feasible domain for x
1) x must be greater than zero
2) The largest x would be when the height is zero.
Then 108 = x2 + 0 and x2 = √108
Thus < x < √108
4) Differentiate
V = 27x - ¼x3
dV = 27 - ¾x2 dx

17. 5) Find critical numbers
dV = 27 - ¾x2 dx
0 = 27 - ¾x2
3x2 = 108
x = ±6
6) Evaluate at endpoints of domain and
at critical numbers
f(0) = 0
f(6) = 108
f(√108) = 0

18. 7) Pick value that gives the largest or smallest value
f(0) = 0
f(6) = 108
f(√108) = 0
maximum
so x = 6 and h = 3
dimensions are 6 x 6 x 3

19. Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.

20. 2) Use equation not being maximized to eliminate a variable
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.
1) Develop 2 equations
2) Use equation not being maximized to eliminate a variable

21. 3) Determine feasible domain for x
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.
3) Determine feasible domain for x
4) Differentiate
5) Find critical numbers

22. 6) Evaluate at endpoints of domain and at critical numbers
Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.
6) Evaluate at endpoints of domain and
at critical numbers
7) Pick value that gives the largest or
smallest value

23. Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.
The numbers are 24 and 8

24. Find the length and width of a rectangle that has an area of 64 and a minimum perimeter.

25. Find the length and width of a rectangle that has an area of 64 and a minimum perimeter.
Area = l . w
64 = l . w
64/w = l
Perimeter = 2l + 2w
Perimeter = 2(64/w) + 2w
y' = -128/w2 + 2
0 = -128/w2 + 2
8 = w
Length = 8 and width = 8

26. Find the point on the graph of the function f(x) = x2 that is closest to the point (2, ½).

27. Find the point on the graph of the function f(x) = x2 that is closest to the point (2, ½).
y = x2
distance = √(2 - x)2 + (½ - y)2
distance = (x4 - 4x )½
distance ' = ½(x4 - 4x )-½ (4x3 - 4)
0 = ½(x4 - 4x )-½ (4x3 - 4)
1 = x

28. A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?

29. A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?
$330

30. Attachments