1
f
(
x
,
y
)
≡
A
x
2
+
B
x
y
+
C
y
2
{\displaystyle f(x,y)\equiv Ax^{2}+Bxy+Cy^{2}}
f
(
x
+
h
,
y
+
k
)
{\displaystyle f(x+h,\ y+k)}
h
{\displaystyle h}
k
{\displaystyle k}
Solution.
∂
f
∂
x
=
2
A
x
+
B
y
,
∂
f
∂
y
=
B
x
+
2
C
y
;
{\displaystyle {\frac {\partial f}{\partial x}}=2Ax+By,\qquad {\frac {\partial f}{\partial y}}=Bx+2Cy;}
∂
2
f
∂
x
2
=
2
A
,
∂
2
f
∂
x
∂
y
=
B
,
∂
2
f
∂
y
2
=
2
C
.
{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}=2A,\qquad {\frac {\partial ^{2}f}{\partial x\partial y}}=B,\qquad {\frac {\partial ^{2}f}{\partial y^{2}}}=2C.}
The third and higher partial derivatives are all zero. Substituting in (67 ) ,
f
(
x
+
h
,
y
+
k
)
≡
A
x
2
+
B
x
y
+
C
y
2
+
(
2
A
x
+
B
y
)
h
+
(
B
x
+
2
C
y
)
k
+
A
h
2
−
+
B
h
k
+
C
k
2
.
{\displaystyle {\begin{aligned}f\left(x+h,y+k\right)\equiv &Ax^{2}+Bxy+Cy^{2}+\left(2Ax+By\right)h+\left(Bx+2Cy\right)k\\&+Ah^{2}-+Bhk+Ck^{2}.\end{aligned}}}
Ans.
2.
f
(
x
,
y
,
z
)
≡
A
x
2
+
B
y
2
+
C
z
2
{\displaystyle f(x,y,z)\equiv Ax^{2}+By^{2}+Cz^{2}}
f
(
x
+
l
,
y
+
m
,
z
+
n
)
{\displaystyle f(x+l,\ y+m,z+n)}
l
{\displaystyle l}
m
{\displaystyle m}
n
{\displaystyle n}
Solution.
∂
f
∂
x
=
2
A
x
,
∂
f
∂
y
=
2
B
y
,
∂
f
∂
z
=
2
C
z
;
{\displaystyle {\frac {\partial f}{\partial x}}=2Ax,\qquad {\frac {\partial f}{\partial y}}=2By,\qquad {\frac {\partial f}{\partial z}}=2Cz;}
∂
2
f
∂
x
2
=
2
A
,
∂
2
f
∂
y
2
=
2
B
,
,
∂
2
f
∂
z
2
=
2
C
,
∂
2
f
∂
x
∂
y
=
∂
2
f
∂
y
∂
z
=
∂
2
f
∂
z
∂
x
=
0.
{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}=2A,\qquad {\frac {\partial ^{2}f}{\partial y^{2}}}=2B,\qquad ,{\frac {\partial ^{2}f}{\partial z^{2}}}=2C,\qquad {\frac {\partial ^{2}f}{\partial x\partial y}}={\frac {\partial ^{2}f}{\partial y\partial z}}={\frac {\partial ^{2}f}{\partial z\partial x}}=0.}
The third and higher partial derivatives are all zero. Substituting in (68 ) ,
f
(
x
+
l
,
y
+
m
,
z
+
n
)
≡
A
x
2
+
B
y
2
+
C
z
2
+
2
A
x
l
+
2
B
y
m
+
2
C
z
n
+
A
l
2
+
B
m
2
+
C
n
2
.
{\displaystyle {\begin{aligned}f\left(x+l,y+m,z+n\right)\equiv &Ax^{2}+By^{2}+Cz^{2}+2Axl+2Bym+2Czn\\&+Al^{2}+Bm^{2}+Cn^{2}.\\\end{aligned}}}
Ans.
3.
f
(
x
,
y
)
≡
x
tan
y
{\displaystyle f(x,y)\equiv {\sqrt {x}}\tan y}
f
(
x
+
h
,
y
+
k
)
{\displaystyle f(x+h,y+k)}
h
{\displaystyle h}
k
{\displaystyle k}
4.
f
(
x
,
y
,
z
)
≡
A
x
2
+
B
y
2
+
C
z
2
+
D
x
y
+
E
y
z
+
F
z
x
{\displaystyle f(x,\ y,\ z)\equiv Ax^{2}+By^{2}+Cz^{2}+Dxy+Eyz+Fzx}
f
(
x
+
h
,
y
+
f
c
,
z
+
l
)
{\displaystyle f(x+h,\ y+fc,\ z+l)}
h
{\displaystyle h}
k
{\displaystyle k}
l
{\displaystyle l}
149. Maxima and minima of functions of two independent variables.
f
(
x
,
y
)
{\displaystyle f(x,y)}
maximum at
x
=
a
,
y
=
b
{\displaystyle x=a,\,y=b}
f
(
a
,
b
)
{\displaystyle f(a,b)}
f
(
x
,
y
)
{\displaystyle f(x,y)}
x
{\displaystyle x}
y
{\displaystyle y}
a
{\displaystyle a}
b
{\displaystyle b}
f
(
a
,
b
)
{\displaystyle f(a,b)}
minimum at
x
=
a
,
y
=
b
{\displaystyle x=a,\,y=b}
f
(
a
,
b
)
{\displaystyle f(a,b)}
f
(
x
,
y
)
{\displaystyle f(x,y)}
x
{\displaystyle x}
y
{\displaystyle y}
a
{\displaystyle a}
b
{\displaystyle b}
These definitions may be stated in analytical form as follows:
If, for all values of
h
{\displaystyle h}
k
{\displaystyle k}
(A )
f
(
a
+
h
,
b
+
k
)
−
f
(
a
,
b
)
=
{\displaystyle f(a+h,b+k)-f(a,b)=}
a negative number , then
f
(
a
,
b
)
{\displaystyle f(a,b)}
maximum value of
f
(
x
,
y
)
{\displaystyle f(x,y)}
If
(B )
f
(
a
+
h
,
b
+
k
)
−
f
(
a
,
b
)
=
{\displaystyle f(a+h,b+k)-f(a,b)=}
a positive number , then
f
(
a
,
b
)
{\displaystyle f(a,b)}
minimum value of
f
(
x
,
y
)
{\displaystyle f(x,y)}
These statements may be interpreted geometrically as follows: a point
P
{\displaystyle P}
z
=
f
(
x
,
y
)
{\displaystyle z=f(x,\ y)}
