To find the extreme values (maximum or minimum) of a function f(x,y) of two variables: 1. Take the partial derivatives fx(x,y) and fy(x,y) and set them equal to 0 to find the stationary points. 2. At each stationary point, calculate the second partial derivatives fxx, fyy, and fxy. 3. Use the signs of fxx and the determinant of the Hessian (fxx*fyy - fxy^2) to determine whether the stationary point is a maximum or minimum. If the determinant is positive and fxx is negative, it is a maximum; if the determinant is positive and fxx is positive, it

2.  Let f(x,y) be a function of two variables x and
y. At x=a, y=b, f(x,y) is said to have maximum
or minimum value, if f(a,b)>f(a+h,b+k) or
f(a,b)<f(a+h,b+k) respectively where h and k
are small values.
Example: The maximum value of
f(x,y)=x3
+3xy2
-3y2
+4 is 36 and minimum value
is -36
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3. Necessary & Sufficient conditions
Extreme value:
 f(a,b) is said to be an extreme value os f, if it is a
maximum or minimum value.
 The necessary conditions for f(x,y) to have a
maximum or minimum at (a,b) are
fx(a,b)=0, fy(a,b)=0.
 Sufficient condition:
i) f(a,b) is a maximum value if ln-m2
>0 and l>0
ii) f(a,b) is a minimum value if ln-m2
<0 and l<0
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4.  Example 1: The extreme values of u=x2
y2
-5x2
-8xy-5y2
are
-8 and -80
 Example 2: The extreme value of x2
+y2
+6x+12 is 3
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5. Stationary point:
 The point (a,b) is said to be a stationary point of f(x,y) if
fx(a,b)=0 and fy(a,b)=0.
 Thus every extreme value is a stationary value but the
converse may not be true.
Example 1: The stationary point of x2
y+xy2
-axy are (0,0)
and (a/3,a/3)
Example 2: The stationary point of x4
+ y4
– 2x2
+ 4xy –
2y2
as x>0 and y>0 are (1.414,-1.414) and (-1.414,1.414).
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6. Working rule:
1. Find ∂f/∂x=0 and ∂f/∂y=0. Solve these equations for
x and y. Let (a,b) and (c,d) etc., are stationary points.
2. Find r= ∂2
f/∂x2,
s= ∂2
f/∂x∂y, t= ∂2
f/∂y2,
for each pair of
values obtained in step 1.
3. (i) if rt-s2
>0 and r<0 at (a,b) then f has a maximum
(ii) if rt-s2
>0 and r>0 at (a,b) then f has a minimum
(iii) if rt-s2
<0 at (a,b) then f has neither maximum
and nor minimum.
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7. Example:
Discuss the maxima and minima of x2
y+xy2
-axy.
Solution: Here we get stationary point (a/3, a/3) and rt-
s2
>0and r>0 at (a/3,a/3)
Thus f(x,y) has minimum at (a/3, a/3) and it is
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