Counter example is with n=1
f(x, y) = x + y.

Now, x + y + z = 120 => z = 120 – x – y
f = xy + yz + zx
f = xy + y(120-x-y) + x(120-x-y) = 120x + 120y – xy – x² – y²
Hence, ^?f⁄_?x = 120 – y – 2x and ^?f⁄_?y = 120 – x – 2y
putting ^?f⁄_?x and ^?f⁄_?y equals to 0 we get, (x, y)=>(40, 40)
Now at (40,40), r=^?2f⁄_?x² = -2 < 0, s = ^?2f⁄_?x?y = -1, and t = ^?2f⁄_?y² = -2
hence, rt – s² = 5 > 0
since, r<0 and rt – s² > 0 f(x,y) has maixum value at (40,40),
Hence, maximum value of f(40,40) = 120 – 40 – 40 = 40,
Hence, x = y = z = 40.

The function takes constant values along a circle(observe function)
But it is composed of arc sine function. Hence, we will have critical points at equal intervals
Hence, re are infinite such points.

Point where function f(x,y) eir have maximum or minimum value is called saddle point. i.e, ^?f⁄_?x = 0 & ^?f⁄_?y = 0.

Taylor?s series expansion is given by
\(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\
+2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)
Thus fifth term is given by \(2 \frac{(x-a)(x-b)}{2!} f_{xy} (a,b)\)..(1) where a=1, b=?/4 & \(f_{xy}=\frac{?}{?x}(\frac{?f(x,y)}{?x}) = \frac{?}{?x}(\frac{?e^x cos?y}{?y}) =- e^x \,sin?y \) at (1, \(\frac{?}{4}), f_{xy}=\frac{-e}{\sqrt{2}}\) substituting in (1)
We get fifth term as \(2 \frac{(x-1)(x-?/4)}{2!} \frac{-e}{\sqrt{2}} = \frac{-e(x-1)(y-\frac{?}{4})}{\sqrt{2}}\).

Differentiating f_xx = -y².sin(xy)
f_yy = -x².sin(xy)
f_xy = -yx.sin(xy)
Observe that f_xx. f_yy – (f_xy)²
Hence, it is a saddle point.

The objective function is f(x,y,z) = x² + y² + z²
compute gradient ∇f = 2x i + 2y j + 2z k
Now compute gradient of function x + y + z = 9
which is
= i + j + k
Using Lagrange condition we have
∇f = λ . ∇_g
2x i + 2y j + 2z k = λ * (i + j + k)
⇒ x = y =z
Put this back into constraint function we get
3x = 9 ⇒ (x,y,z) = (3,3,3).

For function f(x,y) to have minimum value at (a,b)
rt – s²>0 and r>0
where, r = ^?2f⁄_?x², t=^?2f⁄_?y², s=^?2f⁄_?x?y, at (x,y) => (a,b).

In lagrange?s orem of maxima of minima one can not determine nature of stationary points.

Given,f(x,y) = y² + 4xy + 3x² + x³
Now,^?f⁄_?x = 4y + 6x + 3x² and ^?f⁄_?y = 2y + 4x
Putting,^?f⁄_?x and ^?f⁄_?y = 0,and solving two equations,we get,
(x,y) = (0,0) or (2/3, -4/3)
Now,at (0,0) r= ^?2f⁄_?x²=6+6x=6>0 and t= ^?2f⁄_?y² =2>0 and s= ^?2f⁄_?x?y=4
hence, rt – s² = 12 – 16<0,hence it has no extremum at this point.
Now at (²⁄₃,-⁴⁄₃) r=^?2f⁄_?x²= 6 + 6x = 10>0 and t= ^?2f⁄_?y² =2>0 and s= ^?2f⁄_?x?y=4
hence, rt – s² = 20 – 16 > 0 and r>0, hence it has minimum at this point.(²⁄₃, –⁴⁄₃).