1. Expand f(x,y) = xº y about the point (0,0) using Taylor's series formula
2. Find the maximum and minimum values of the function f(x,y) = x + y - 3axy
3. Expand tan at the point (1,1) by Taylor's series as far as quadratic terms
4. Expand in the Taylor's series e' log (1 + y) in powers of x and y up to the third degree
5. Examine the function f(x,y) = x'y? (12 - x - y) for extreme values
6. Examinef (x,y) = x + y3 - 12x - 3y + 20 for its extreme values
7. Expand in the Taylor's series e'log (1 + y) in powers of x and y up to the third degree
Answers
1) TAYLOR POLYNOMIALS AND TAYLOR SERIES
The following notes are based in part on material developed by Dr. Ken Bube of the
University of Washington Department of Mathematics in the Spring, 2005.
1 Taylor Polynomials
The tangent line to the graph of y = f(x) at the point x = a is the line going through the
point ( ) a, f (a) that has slope f '(a) . By the point-slope form of the equation of a line, its
equation is
y f (a) = f '(a)(x a)
y = f (a) + f '(a)(x a)
As you have seen in Math 124, the tangent line is a very good approximation to y = f(x)
near x = a, as shown in Figure 1.
FIGURE 1. The line y = f (a) + f '(a)( ) x a tangent to the graph of y = f(x)
2) What are the maxima and minima of x3+y3−3axy?
Let f(x,y)=x3+y3−3axy.
At the critical values of the function, both the first derivatives are zero.
⇒fx=3x2−3ay=0 and fy=3y2−3ax=0.
⇒y=x2a⇒3(x2a)2−3ax=0.
⇒x4a2−ax=0⇒x4−a3x=0.
⇒x(x3−a3)=0.
⇒x=0 or x=a.
When x=0,y=0 and when x=a,y=a.
fxx=6x,fyy=6y and fxy=−3a.
At (x,y)=(0,0),
fxxfyy−f2xy=36xy−9a2=−9a2<0.
⇒(x,y)=(0,0) is a saddle point.
At (x,y)=(a,a),
fxxfyy−f2xy=36xy−9a2=27a2>0
Further, fxx=6a>0 if a>0 and fxx=6a<0 if a<0.
⇒(x,y)=(a,a) represents a minimum if a>0 and represents a maximum if a<0.
When (x,y)=(a,a),f(x,y)=x3+y3−3axy=−a3.
If a=0, the function becomes x3+y3, which does not have any maximum or minimum and has a saddle point at (x,y)=(0,0).
Therefore, we conclude as under:
If a>0, the function does not have a maximum but has a local minimum at (a,a) having a value −a3.
If a<0, the function does not have a minimum but has a local maximum at (a,a) having a value −a3.
If a=0, the function does not have either a maximum or a minimum.
3) Refer the 1 st two attachments..⤴️
4) e^x log (1+y) = y +1/2 (2x-y-y^2)+1/6(3x^2y-3xy^2+2y^3)
Given: f(x,y) = xº y about the point (0,0)
To Find: maximum and minimum values of the function
Solution:
By Taylor Series Formula,
The tangent line to the graph of y = f(x) at the point x = a
f (a) that has slope f '(a)
By the point-slope form, the equation is
y = f (a) + f '(a)(x a)
the tangent line is a approximation to y = f(x)
near x = a
Therefore, the line y = f (a) + f '(a)(x) is a tangent to the graph of y = f(x)
For the maxima and minima of x+y−3axy
Let f(x,y)=x+y−3axy.
At the critical values of the function, both the first derivatives are zero.
This means that f(x) =3x−3ay=0
f(y) =3y−3ax=0.
y=x2a = 3(x2a)2−3ax=0.
x4a2−ax=0
=x4−a3x=0.
=x(x3−a3)=0.
x=0 or x=a.
When x=0 and y=0
x=a and y=a.
fxx=6x and
fyy=6y and
fxy=−3a.
At (x,y)=(0,0),
fxxfyy−f2xy=36xy−9a2=−9a2<0.
(x,y)=(0,0) is a critical point.
At (x,y)=(a,a),
fxxfyy−f2xy=36xy−9a2=27a2>0
Further, fxx=6a>0 if a>0 and fxx=6a<0 if a<0.
(x,y)=(a,a) represents a minimum if a>0 and represents a maximum if a<0.
When (x,y)=(a,a),f(x,y)=x3+y3−3axy=−a3.
If a=0, the function becomes x3+y3, which does not have any maximum or minimum and has a critical point at (x,y)=(0,0).
Therefore,
If a>0, the function does not have a maximum but has a local minimum at (a, a)
If a<0, the function does not have a minimum but has a local maximum at (a, a)
If a=0, then the function does not have either a maximum or a minimum.
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