p.372 #1
Maximize area:
Constraint:
Increases for 
Decreases for 
maximum at 
Total width: 150 ft
Length: 100 ft
p.372 #2
Minimize wall length:
Constraint:
Decreases for 
Increases for 
minimum at 
a) Rooms should be 14.289 ft by 24.495 ft
b) 10 rooms: 
p.373 #3
Maximize area:
Constraints:
a) smallest when 
so 
largest when using all fence except minimum used by square
(40 ft) so 
b)
[0, 100] x [0, 20000]
c)
Decreases for 
Increases for 
minimum at 
Looking for maximum, so consider endpoints:
maximum area is 17,522.222 sq. ft.
p.372 #4
Constraints:
Maximum circle size
Decreases for 
Increases for 
minimum at 
b) maximum must occur at endpoint
maximum area when 
(all fence used on
circle)
p.372 #5
Maximize volume:
Constraint:
Increases for 
Decreases for 
maximum at 
a) 6.325 cm by 6.325 cm by 3.162 cm
b) conjecture: Depth is half of width
p.372 #7
Minimize cost:
Constraint:
Decreases for 
Increases for 
minimum at 
p.372 #10
Minimize area:
Constraints:
Increasing for 
Decreasing for 
maximum at 
Minimum must occur at endpoint
[0, 150] x [0, 15000]
Smallest r is 20.
Largest r occurs when
least straight lengths are used (i.e. ).
minimum occurs at 
p.372 #11
Minimize length of ladder, l:
Relate x and y using similar triangles:
Re-write l in one variable:
Find minimum using derivative:
(minimum 
is equivalent to
minimum l)
*YUCK! TIME TO USE MY CALCULATOR!*
Graph :
[0, 12] x [0, 200]
Minimum occurs at (5, 125)
Since minimum is 125, minimum l is approximately 11.180 feet.
p.372 #13
Maximize
volume of cylinder:
Relate r and h using given perimeter:
Re-write V in one variable:
Find maximum using derivative:
Max occurs when 
, so 
400 mm radius and 200 mm height
p.372 #15
a)
b)
Relate r and h:
Rewrite A in terms of r:
c) Find minimum using derivative:
4.133 cm radius, 8.266 cm height
Can is short and fat
Ratio of diameter to altitude is 1 to 1
d) 
The normal can uses close to the same amount of metal
Normal can uses about 1.5% more metal
e) 
about $6.4 million