5.2 first and second derivative test

1. Applied Maximum and
Minimum Problems

2. Optimization Problems
1.Draw an appropriate figure and label the quantities
relevant to the problem.
2.Write a primary equationthat relates the given and
unknown quantities.
3.If necessary, reduce the primary equation to 1 variable
(use a secondary equation if necessary).
4.Determine the desired max/min using the derivative(s).
5.Check solutions with possible values (domain).

3. Example:An open box is to be made from a 16” by 30” piece of
cardboard by cutting out squares of equal size from the four
corners and bending up the sides. What size should the squares
be to obtain a box with the largest volume?V=l×w×h l=30-2x w=16-2x h=x V=30-2x( )×16-2x( )×x V=30x-2x
2
( )×16-2x( )=480x-60x
2
-32x
2
+4x
3 V=4x
3
-92x
2
+480x V'=12x
2
-184x+480=0 V'=43x
2
-46x+120( )=0 V'=4x-12( )3x-10( )=0
x
x
16
300£x£8 x=
10
3
andx=12 x=
10
3

4. Example:An offshore oil well located at a point W that is 5
km from the closest point A on a straight shoreline. Oil is to
be piped from W to a shore point B that is 8 km from A by
piping it on a straight line under water from W to some shore
point P between A and B and then on to B via pipe along the
shoreline. If the cost of laying pipe is $1 million under water
and $½ million over land, where should the point P be
located to minimize the coast of laying the pipe?
5 km
A P B
x 8 –x
8 km
WWP=x
2
+25 PB=8-x C=1×x
2
+25+
1
2
×8-x() 0£x£8 C'=
1
2x
2
+25
×2x-
1
2
=0 x
x
2
+25
=
1
2 x
2
+25=2x x
2
+25=4x
2 3x
2
=25 x=
5
3
x=-
5
3 C
5
3
æ
è
ç
ö
ø
÷»8.330127
C(8)»9.433981
abs max

5. Example: You have 200 feet of fencing to enclose two adjacent
rectangular corrals. What dimensions should be used to maximize the
area?A=2x×
200-4x
3
=
400x-8x
2
3 P=4x+3y=200 y=
200-4x
3 A'=
3400-16x( )
9
=0 x=25 0£x£200 y=
100
3

6. Example:Find the radius and height of the right circular
cylinder of largest volume that can be inscribed in a right
circular cone with radius 4 inches and height 10 inches.
r
hV=pr
2
h
10-h10-h
r
=
10
4 h=10-
5
2
r V=pr
2
10-
5
2
r
æ
è
ç
ö
ø
÷ V=10pr
2
-
5p
2
r
3 V'=20pr-
15p
2
r
2
=0 5pr4-
3
2
r
æ
è
ç
ö
ø
÷=0 r=
8
3 0£r£4 r(0)=0
r
8
3
æ
è
ç
ö
ø
÷»74.42962
r(4)=0
abs maxh=
10
3