Applying this to (F ) ,
A
=
∂
2
f
∂
x
2
,
B
=
∂
2
f
∂
y
2
,
C
=
∂
2
f
∂
x
∂
y
{\displaystyle A={\tfrac {\partial ^{2}f}{\partial x^{2}}},\quad B={\tfrac {\partial ^{2}f}{\partial y^{2}}},\quad C={\tfrac {\partial ^{2}f}{\partial x\partial y}}}
(F ) , and therefore also the left-hand member of (E ) , has the same sign as
∂
2
f
∂
x
2
{\displaystyle {\tfrac {\partial ^{2}f}{\partial x^{2}}}}
∂
2
f
∂
y
2
{\displaystyle {\tfrac {\partial ^{2}f}{\partial y^{2}}}}
∂
2
f
∂
x
2
∂
2
f
∂
y
2
−
(
∂
2
f
∂
x
∂
y
)
2
>
0.
{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial x\partial y}}\right)^{2}>0.}
Hence the following rule for finding maximum and minimum values of a function
f
(
x
,
y
)
{\displaystyle f(x,y)}
First Step . Solve the simultaneous equations
∂
f
∂
x
=
0
,
∂
f
∂
y
=
0.
{\displaystyle {\frac {\partial f}{\partial x}}=0,\qquad {\frac {\partial f}{\partial y}}=0.}
Second Step . Calculate for these values of x and y the value of
Δ
=
∂
2
f
∂
x
2
∂
2
f
∂
y
2
−
(
∂
2
f
∂
x
∂
y
)
2
.
{\displaystyle \Delta ={\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial x\partial y}}\right)^{2}.}
Third Step. The function will have a
maximum if
Δ
>
0
{\displaystyle \Delta >0}
and
∂
2
f
∂
x
2
(
or
∂
2
f
∂
y
2
)
<
0
;
{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}\left({\text{or}}{\frac {\partial ^{2}f}{\partial y^{2}}}\right)<0;}
minimum if
Δ
>
0
{\displaystyle \Delta >0}
and
∂
2
f
∂
x
2
(
or
∂
2
f
∂
y
2
)
>
0
;
{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}\left({\text{or}}{\frac {\partial ^{2}f}{\partial y^{2}}}\right)>0;}
neither a maximum nor a minimum if
Δ
<
0
{\displaystyle \Delta <0}
.
The question is undecided if
Δ
=
0
{\displaystyle \Delta =0}
.[ 1]
The student should notice that this rule does not necessarily give all maximum and minimum values. For a pair of values of
x
{\displaystyle x}
y
{\displaystyle y}
First Step may cause
Δ
{\displaystyle \Delta }
The question of maxima and minima of functions of three or more independent variables must be left to more advanced treatises.
Illustrative Example 1.
3
a
x
y
−
x
3
−
y
3
{\displaystyle 3axy-x^{3}-y^{3}}
Solution.
f
(
x
,
y
)
=
3
a
x
y
−
x
3
−
y
3
.
{\displaystyle f(x,\ y)=3axy-x^{3}-y^{3}.}
First step.
∂
f
∂
x
=
3
a
y
−
e
x
2
=
0
,
∂
f
∂
y
=
3
a
x
−
3
y
2
=
0.
{\displaystyle {\frac {\partial f}{\partial x}}=3ay-ex^{2}=0,\qquad {\frac {\partial f}{\partial y}}=3ax-3y^{2}=0.}
Solving these two simultaneous equations, we get
x
=
0
,
x
=
a
,
{\displaystyle x=0,\qquad x=a,}
y
=
0
;
y
=
a
.
{\displaystyle y=0;\qquad y=a.}
↑ The discussion of the text merely renders the given rule plausible. The student should observe that the case
δ
=
0
{\displaystyle \delta =0}
