3.7 Optimization Problems. After this lesson, you should be able to: Be able to model the real-life...
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Transcript of 3.7 Optimization Problems. After this lesson, you should be able to: Be able to model the real-life...
3.7 Optimization Problems
After this lesson, you should be able to:
Be able to model the real-life problem to mathematical problem.Solve applied maximum and minimum problems.
One of the most valuable aspects of differential calculus is the ability to find where something is maximized or minimized. When you think about your own personal finances, aren't two of the most important questions you can ask "When/where do I have the least amount of cash and when/where do I have the most?" What factors led to that? At what point is this occurring and why?
Optimization Problems
Manufacturers want to maximize their profit while minimizing their costs. Thanks to calculus, we've learned that our derivative tells us plenty about our function. And when our derivative is zero, WE WANT TO PAY CLOSE ATTENTION TO WHAT IS HAPPENING TO OUR FUNCTION. Remember that max's and min's can occur where our derivative is zero or undefined, so applying this concept is not too difficult. Now, let's define the steps in analyzing optimization problems.
Optimization Problems
Example
Example 1 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?
hxhxV 2),(
Solution
Let the length of the base be x inch,
xx
hand the height be h inch.
Then the volume of the box in terms of x and h inch is
Primary EquationNotice that “surface area of 108
in2 ” provides 1084),( 2 xhxhxS Secondary Equation
Example
Example 1 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?
xxh 4/)108( 2
Solution
From the secondary equation, we have
xx
h
Then the volume of the box in terms of x in the primary equation is
Now, before we aim on V(x) and try to find its maximum volume while x has certain length, we should find the feasible domain of x.
4/274/)108()( 322 xxxxxxV
Notice that 22 4108),( xxhxhxS
This means 1080 x
0xand
Then taking the derivative of V(x) and set to zero. We have 04/327)(' 2 xxV
6xThe critical number within the feasible domain is
Feasible Domain
Compute the endpoints 0)0( V
and
108)6( V
0)108( Vand Therefore, V(6, 3) is maximum when x = 6
and h = 3. The box dimension is 6 x 6 x 3.
324/)36108(4/)108( 2 xxh
Guidelines for Solving Applied Minimum and Maximum
Problems
Example
Example 2 You are designing an open box to be made of cardboard that is 10 inches by 15 inches. The box will be formed by making the cuts at each corner. How long should you make the cuts? What is the maximum volume?
15
10
Example
x215
Solution
Let the side of square to be cut be x inch.
Then the dimensions of the boxd can be interpreted asLength =Therefore, the volume is
xxxxV )210)(215()(
15
10
xx
xx
xx
xx
xxx 150504 23
x210 Width = xHeight =
Notice that 102 x
Feasible Domain
0x and
50 xThen
Then taking the 1st derivative of V(x) and set to zero. We have
015010012)(' 2 xxxVThe critical number within the feasible domain is
xxxxV 150504)( 23
Example
6
7525 x
domain)
feasible thein not is 56
7525(
x
27
)7710(125
6
7525
V
Then taking the 2nd derivative of V(x), we have 10024)('' xxV
Therefore, is maximum
volume when we cut 4 squares of side .
15010012)(' 2 xxxV
Example
07201006
752524
6
7525''
V
27
)7710(125
6
7525
V
6
7525
Example
Example 3 Which points on the graph of y = 4 – x2 are closest to the point (0, 2)?
x
y
d (x, y)
Solution
The distance between the point (0, 2) to any point (x, y) on the graph is 2222 )2()2()0( yxyxd
Notice that y = 4 – x2, then
43)24()( 24222 xxxxxd
Let D(x) = d2(x), then43)( 24 xxxD
Then taking the 1st derivative of D(x) and set to zero. We have
0)32(264)(' 23 xxxxxDThe critical number within the feasible domain is
Example
0x and 2
6x
Then taking the 2nd derivative of D(x), we have
060'' D
)12(6612)('' 22 xxxD
0212
32
2
6''
Dan
d This verifies that x = 0 yields a relative maximum andyield a minimum distance to the given point.
2
6x
x
y
Notice that in this example, we can not find the absolute maximum, but only the relative maximum. However, we do have absolute minimum.
43)( 24 xxxD
Example
Example 4 A man is in a boat 2 miles from the nearest point R on the coast. He is trying to go to a point Q, located 3 miles down the coast and 1 mile inland. He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time?
Q
2
1
3
Solution
Let the point on the coast toward by the man be P, and PR = x, then PS = 3 – x. Then
PR Sx
3 – x
M
42 xMP
1)3( 2 xPQ
Example
Q
2
1
3Then the total time from point M to Q will be
PR Sx
3 – x
M
4
106
2
4)(
22
xxxxT
01064
3
42 22
xx
x
x
x
dx
dT
Then taking the 1st derivative of T(x) and set to zero. We have
)106(4
96
4 2
2
2
2
xx
xx
x
x012896 234 xxxx
The feasible domain for x is [0, 3]. The solution for x within the domain is x = 1.
x
y
Then compute
Example
4
101)0( T
4
53)1( T
4
1321)3(
T
We conclude that when x = 1 yields the minimum time.
Example
Example 5 Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum area?Solution
r
x
x
Let the side of the square be x and the radius of the circle be r. Then
Since the total length of wire is 4 feet, then
22),( rxrxA Primary Equation
Secondary Equation
424 rx So,
)1(2 xr
Example
r
x
x
The feasible domain of x is restricted by the square’s perimeter
22 )1(2
)(
xxxA
)1(2 x
r
22 )1(4 xx
]48)4[(1 2 xx
10 x
08)4(2
x
dx
dATaking the 1st derivative of A(x) and set to zero. We have
The critical number within the feasible domain is
4
4
x
Then compute
4
)0( A
4
4)4
4(
A
1)1( A
We conclude that when yields the maximum area. That means all the wire is used to form circle.
Example
0x
x
y
Example
Example 6 A square fountain has a water pool of d feet wide around. A repairman tried to fix a water pipe in the fountain. He used 2 pieces of wood of length L(<d) to access the water pipe in the center. What is the minimum length of the wood board? How did he place these 2 pieces of wood?SolutionWe simplify the problem to the following question: in a square of side length of d, find the 2 perpendicular segments of length L and when L can reach its minimum.
d
dLL
Example
Let the acute angle between one wood board and outer edge of the fountain be θ. Extending TS to R.
d
dL
L
Q
S
T
R P
Then in TRQ, RQ = d tan And PQ = d – Lcos
Therefore RP = RQ – PQ = d tan – (d – Lcos )
= d (tan – 1) + Lcos In PRS, RS = RP sin =[d (tan – 1) + Lcos ]sin In TRQ, RS + ST = d/cos Or, [d (tan – 1) + Lcos ]sin + L = d/cos Solve for L, we get
cossin1
cossin)(
dL
ExampleNow we seek what value will yields the minimum L. Or,We only need what value will yields the minimum value to f( ). We notice that 0 < < /2 is the feasible domain.
)(cossin1
cossin)(
dfdL
Taking the 1st derivative of f( ), we have
2
2
)cossin1(
])cos(sincossin1)[sin(cos
2)cossin1(
2cos)cos(sin)cossin1)(sin(cos)('
f
2)cossin1(
)cossin)(sin(cos
2)cossin1(
)cos)(sin2)(sin2/1(
The critical number within the feasible domain is
4
0( 2
and are not in the 0 < < /2 )
ExampleWe are going to take the 2nd derivative of f( ). The derivative of the numerator is
4
2
)cossin1(
)cossin1)(1cossin3)(cos(sin)(''
f
2)cossin1(
)cos)(sin2)(sin2/1()('
f
d
d )]cos(sin2sin2/1[
)cos(sincossin)cos(sin2cos ]cossin)cos(sin)[cos(sin 2
)1cossin3)(cos(sin
4)cossin1(
2cos)cossin1(2)cos)(sin2)(sin2/1(
ExampleWe simplify the 2nd derivative of f( ). We get
4
2
)cossin1(
)cossin1)(1cossin3)(cos(sin)(''
f
2)cossin1(
)cos)(sin2)(sin2/1()('
f
4)cossin1(
2cos)cossin1(2)cos)(sin2)(sin2/1(
4
2
)cossin1(
)cossin1)(1cossin3)(cos(sin
4)cossin1(
)cossin1)(cos(sin2cos2sin
4
2
)cossin1(
)cossin1()sin)(coscos(sin2sin
Example
4)cossin1(
)cossin1)(cos(sin)(''
f
])sin(cos2sin)cossin1)(1cossin3[( 2
4)cossin1(
)cossin1)(cos(sin
)2sin1(2sin2sin
2
1112sin
2
3
When =/4, we know that all the factors are positive. So 0)
4(''
f
When =/4, f( ) yields the minimum value. Therefore, ddfL
3
22)4()(
)(cossin1
cossin)(
dfdL
x
y
Homework
Pg. 223 3-9 odd, 17 – 23 odd, 27, 29, 33, 49