Transcript Chapter 14 - SaigonTech

Lial/Hungerford/Holcomb/Mullins:
Mathematics with Applications 11e
Finite Mathematics with Applications 11e
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Chapter 14

Multivariate Calculus
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Section 14.1

Functions of Several Variables
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Example:
9
Let f ( x, y )  4 x  2 xy  , and find each of the given
y
quantities.
2
(a) f (1,3)
Solution:
Replace x with −1 and y with 3:
f (1, 3)  4(1)2  2(1)(3) 
9
 4  6  3  1.
3
(b) The domain of f
Solution:
Because of the quotient 9/y, it is not possible to replace y with zero.
So, the domain of the function f consists of all ordered pairs ( x, y)
such that y  0.
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A saddle
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Graph of a function of 2 variables
Find the domain and the range of
Domain:
Range
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Exercises: Find the domain and the range of functions
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Section 14.2

Partial Derivatives
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Example:
Solution:
Let f ( x, y)  ln( x 2  y). Find fx and fy.
Recall the formula for the derivative of the natural logarithmic
function. If y  ln( g ( x)), then y  g( x) / g ( x).
Using this formula and treating y as a constant, we obtain
Similarly, treating x as a constant leads to the following result:
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Marginal Productivity
A company that manufactures computers has
determined that its production function is given
by:
where x is the size of labor force (in work hours
per week) and y is the amount of capital invested
(in units of $1000 ). Find the marginal
productivity of labor and the marginal
productivity of capital when x = 50 and y = 20,
and interpret the results.
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Surface Area of Human
The surface area of human (in m2) is
approximated by: A(M, H) = .202M.425H.725
where M is the mass of the person (in kg) and H
is the height (in meters). Find the approximate
change in surface area under the given condition:
(a) The mass changes from 72kg to 73kg, while
the height remains 1.8m
0.0112
(b) The mass remains stable at 70kg, while the
height changs from 1.6m to 1.7m. 0.0783
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Show that the function z = 5xy satisfies
Laplace’s equation
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Section 14.3

Extrema of Functions of Several
Variables
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Example:
Find all critical points for
Solution:
Since the partial derivatives always exist, we must find all points
(a, b) such that f x (a, b)  0 and f y (a, b)  0.
f ( x, y)  6 x2  6 y 2  6 xy  36 x  54 y  5.
Here,
f x  12 x  6 y  36
and
f y  12 y  6 x  54.
Set each of these two partial derivatives equal to 0:
12 x  6 y  36  0
and
12 y  6 x  54  0.
These two equations form a system of linear equations that we can
rewrite as
12 x  6 y  36
6 x  12 y  54.
To solve this system by elimination, multiply the first equation by
−2 and then add the equations.
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Example:
Find all critical points for
f ( x, y)  6 x 2  6 y 2  6 xy  36 x  54 y  5.
Solution:
Substituting x  7 in the first equation of the system, we have
Therefore, (−7, 8) is the solution of the system.
Since this is the only solution, (−7, 8) is the only critical point for
the given function. By the previous theorem, if the function has a
local extremum, it must occur at (−7, 8).
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≥3
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fx(0, 0) and fy(0, 0) are unfedined
1
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Saddle point
fx(x, y) = -2x
fx(0, 0) = 0
fy(x, y) = 2y
fy(0, 0) = 0
Around (0,0) the function takes
negative values along the x-axis and
positive values along the y-axis.
The point (0, 0) can not be a local extremum.
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The test fails
≥0
Every point on the x-axis and y-axis yields local (and
global) minimum.
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Examine the function for local extrema and saddle point.
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Section 14.4

Lagrange Multipliers
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Example 1:
Solution:
Use Lagrange’s method to find the minimum value of
f ( x, y)  x2  y 2 , subject to the constraint x  y  4.
First, rewrite the constraint in the form g ( x, y )  0 :
x  y  4  0.
Then, follow the steps in the preceding slide.
Step 1
As we saw previously, the Lagrange function is
Step 2
Find the partial derivatives of F:
Step 3
Set each partial derivative equal to 0 and solve the resulting
system:
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Example:
Use Lagrange’s method to find the minimum value of
Solution:
Since this is a system of linear equations in x, y, and λ, it could be
solved by the matrix techniques. However, we shall use a different
technique, one that can be used even when the equations of the
system are not all linear. Begin by solving the first two equations
for λ:
f ( x, y)  x2  y 2 , subject to the constraint x  y  4.
Set the two expressions for λ equal to obtain
Now make the substitution y = x in the third equation of the original
system:
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Example:
Use Lagrange’s method to find the minimum value of
Solution:
Since y  x and   2 x we see that the only solution of the original
system is
f ( x, y)  x2  y 2 , subject to the constraint x  y  4.
Graphical considerations show that the original problem has a
solution, so we conclude that the minimum value of f ( x, y)  x 2  y 2 ,
subject to the constraint x  y  4, occurs when x  2 and y  2.
The minimum value is
f (2,2)  22  22  8.
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Example 2:
Find a rectangle of maximum area that is inscribed in
the elipse
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Example 3:
A builder plans to construct a 3-story building with a
rectangular floor plan. The cost of the building is given
by: xy + 30x + 20y + 474000, where x and y are the
length and thw width of the rectangular floors. What
length and width should be used if the building is to cost
$500000 and have maximum area on each floor?
x = 113.17, y = 169.75, A = 19211 ft2
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Example 4:
Find three positive numbers x, y and z whose sum is 50
and such that xyz2 is as large as possible.
x = 12.5, y = 12.5, z = 25, xyz2 = 97656.25
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Example 5:
Find the dimensions of the closed rectangular box of
maximum volume that can be constructed from 6 ft2 of
material.
x = 1, y = 1, z = 1, Volume = 1 ft3
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Example 6:
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Exercises
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Exercises
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