Practical Problems慕课视频播放-微积分1（Calculus I）-MOOC慕课视频教程-柠檬大学

Today we study the Calculus (I)

The topic is Practical Problems

In practical life

there are many practical optimization problems

how to deal with

the kind of practical problems

The basic idea

is step by step method

The practical problem is changed into

mathematical question

How to do

Now look at the solving method

the concrete Computation Steps

Step 1

draw a picture

and assign appropriate variables

Step 2

write a formula

for our objective function

Write it as Q

And then

express Q as a function

of a single variable

Pay attention to a single variable

Step 4

try to find the critical points

Do you remember

what are our critical points

Yes

there are 3 kinds of points

end points

stationary points

and singular points

And the final step

try to determine

the maximum or minimum

Pay attention to

if in some practical questions

there is a unique stationary point

about the objective function

The corresponding function value at this point

is our desirable maximum or minimum

How to apply the following steps

So now look at Example 1

Try to find the greatest volume

that a right circular cylinder can have

if it is inscribed in a sphere of radius r

So now look at the picture

It is our the geometrical picture

So let the corresponding

height of cylinder be double h

volume be capital V

Look at the corresponding picture

This is R radius

This is our height double h

This is our little r

corresponding radius

So now look at

what is our objective function

Our objective function means

how to get the volume

So volume equals

basic area time the corresponding height

Basic area equals π r square time

where is our height

yes double h

so times double h

It is our V

How to do

Try to become

the objective function

is a function of one variable

By little r square plus h square

equals capital R square

we can get the relationship of h and R

So V equals double π

double π

Where is our little r square

Equals capital R square minus h square

So now our objective function

is a function of one variable

And then

pay attention to the domain h

h between 0 and capital R

Try to find the maximum point

What is it

Try to find the critical point

How to do

Yes

V is a function of h

So V prime for h

equals double π is a constant

The first term

R square h prime equals R square

The second term h cube prime

equals minus 3 h square

Corresponding

try to find

it is our

the V prime for h

Try to find the critical points

Let V prime h equals 0

obtain two points

One is positive

capital R over square root of 3

another is negative value

Pay attention to

delete the negative value

Why

Because h is our height

should be positive number

So negative value is deleted

So now

unique stationary point

corresponding

is our maximum volume

the point

So we can get

the unique stationary point h is

capital R over square root of 3

is our desirable point

is our maximum value point

Corresponding

maximum volume equals

V equals

double π time R square minus little h square

It is our final value

Pay attention to Example 1

How to apply the 5 steps

Write our objective function

of one single variable

Try to find

the stationary point means

ome critical points

and then get the maximum volume

In the similar idea

now look at Example 2

The curved triangle region

bounded by y equals 0

x equals 8

two straight lines

And another curve

y equals x square

Try to find one point P

The P lies in y equals x square

such that

the corresponding

triangle area bounded by

the tangent line through P

y equals 0

and x equals 8

is our the maximum value

So now how to write it

Look at the picture

In this picture

the pink curve

means y equals x square

The green straight line

means x equals 8

This is our

y equals 0

Together

passing P

do some tangent line

capital PT

So now look at

the corresponding triangle region ABC

Try to find the maximum area

So now

as shown in the picture

let the tangent point P(x0, y0)

We try to find

the concrete values for the fixed point

How to write it

Look at the tangent line

The tangent line capital PT

according to the point slope form

What is our point

Yes

P(x0, y0)

How about slope

Yes

slope means the slope value

So we can write it

y minus y0 equals k

k means slope

times x minus x0

What is our slope

Pay attention to P

that's on the pink curve

How about the pink curve

Y equals x square

corresponding slope means double x

taking the value means double x0

So it is our the tangent line

So now try to get

the P lying in y equals x square

So y0 equals x0 square

satisfying this equation

Together we can get

corresponding intersection point

Corresponding A

half x0,0

Corresponding C

8, 0

Corresponding B

8, corresponding 16 x0 minus x0 square

So we can make up the corresponding triangle region

How about the area

Yes

the area equals half time basis and height

And then

we can get

the area of triangle ABC equals half

this is our basis

this is our the corresponding height

Together

write it this form

Yes

it is our objective function

It is a function of one variable

means one single variable

Pay attention to the domain

between 0 and 8

And then

S prime for variable

What is our variable

Yes x0

So prime x0 equals this equation

We can draw it by ourselves

Yes

so now look at this form

S prime x0 let

corresponding value equals 0

Try to find the critical points

So we can obtain

one point x0 equals 16 over 3

and then equals 16

Look at the domain

x0 between 0 and 8

So this value is deleted

There is one and only one stationary point

means the unique stationary point

In practical problem

the corresponding maximum area is

obtained at the stationary point

at the unique stationary point

So the point P is the final result

Corresponding

the maximum area S is the value

is our desirable maximum area

So in Example 2

We using the similar idea by the 5 steps

So now look at our summary

for our practical problems

How to do

Yes

the following steps

to solve the practical optimization problems

How about 5 steps

Look at

do you remember Step 1 until Step 5

Step 1 yes

draw a picture

and assign appropriate variables

May be height

may be length

may be width

means assign appropriate variables

And then

write a formula for the objective function Q

What is objective function

Yes

you find the maximum area

Area is our objective function

The minimum volume

corresponding volume is our objective function

And then try to

express Q as a function of

a single variable

Maybe there are two unknown variables

Try to

make the function into one single variable

And then try to find

the critical points

How many points

Yes

three kinds of points

end points

stationary points

and singular points

he last step

determine the maximum or minimum

If there are more than two points

how to do

Compare the values

So now look at

at the following question and answer

The question is

a corresponding rectangular box

is to be made from

a piece of cardboard 24 inches long

and 9 inches wide

by cutting out identical squares

Pay attention to

identical squares

from the four corners and turning up the sides

Look at our question

There are two questions

Question 1

find the dimension of the box of the

maximum volume

Question 2

what is the corresponding volume

Means give me

how about the value

We have the following idea

Let x be the width of the square

to be cut out

and capital V

the volume of the resulting box

How to write it

Do you remember the basic idea

Write our corresponding objective function

So in this question

V is our objective function

Do you remember the basic formula

V equals the length time width time height

In this question

the original length is 24

Minus from the 4 squares

So double x

So the length is 24 minus double x

Corresponding width

width

original is 9

9 minus double x

So width is 9 minus double x

The height doesn't change

Corresponding volume equals

length times width times height

Together equals

this expression

216 x minus 66 x square plus 4 x cube

It is a polynomial of x

Look at the variable x

It is a function of one variable x

x between 0 and 4 point 5

means the given domain

How to find

Try to find the critical points

So V prime for the variable x

V prime x equals 0

We can obtain two points

One point x equals 2

the other point x equals 9

Compare the domain x between 0 and 4 point 5

So the 9 is deleted

So there are only 3 critical points

How many

Three

how to get it

Two end points 0 and 4.5

and have another stationary point 2

Together

0 2 and 4 point 5

3 critical points

determine which value is maximum

So we can get

corresponding V(0)

V(0) means 0 taking 0 0 0

together volume equals 0

The value V at the point 4 point 5

Taking the value

equals 0

V the value at the point 2

Taking the stationary point 2

The final result equals 200

Now compare the three values 0

0 0 200

Of course 200 is the maximum

So corresponding

x equals 2 is the final result

Corresponding volume is 200

The box is 20 inches long

5 inches wide

and 2 inches deep

Corresponding

3 dimensional

So now the class is over

See you next time

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Practical Problems课程教案、知识点、字幕

微积分1（Calculus I）课程列表：

Course Introduction

Chapter 1 Limits

Homework 1

Chapter 2 The Derivative

Homework 2

Chapter 3 Applications of the Derivative

Test 1

Chapter 4 The Definite Integral

Homework 4

Chapter 5 Applications of the Integral

Homework 5

Chapter 6 Transcendental and Functions

Chapter 7 Techniques of Integration

Homework 7

Chapter 8 Indeterminate Forms and Improper Integrals

Test 2

Practical Problems笔记与讨论

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