Rewrite function as
f(x, y) = (x + y)² + (x + y)³
Put x + y = 1
= 2.

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).

f(x,y) = c is a level curve over which function has constant value
Hence, we have answer as a constant.

Find
f_x = 2x
f_y = 2y
The critical point is
x=0
y=0
(0,0) is critical point
Put it back into function we get
z = 0 + 0 + 199 = 199 is required minimum value.

Taylor?s orem helps in expanding a function into infinite terms however, it can be applied to functions that can be expressed finitely.

Differentiating f1(g1(x), g2(y)) with respect to x and y separately we get
d_x = f1x g1x (x)
d_y = f1y g1y (y)
This implies
g1x = 0
g1y = 0
Which are also set of critical points of f1(x, y)
Thus we have relation as s(1) ∩ s(2) ≠ 0.

First consider se functions as infinite dimension vectors. Given constraint dimension is always less than objective we can apply Lagrange condition. We now have
\(\frac{1}{1}i_1+\frac{1}{2}i_2+\frac{1}{4}i_3+…..\infty=\lambda.(2x_1i_1+2x_2i_2+2x_3i_3+…\infty)\)
Treating equations individually we get
\(x_1=\frac{1}{2\lambda}\)
\(x_2=\frac{1}{4\lambda}\)
and so on.
Putting back in constraint we get
\(1=\frac{1}{(2\lambda)^2}+\frac{1}{(4\lambda)^2}+….\infty\)
\(1=\frac{1}{(\lambda)^2}+(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}….\infty)\)
\(\Rightarrow \lambda = \frac{1}{\sqrt{3}}\)
Hence, we get extreme value (after putting back values of variables in function) as
extreme value = \(\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt[3]{3}}+\frac{1}{2\sqrt[5]{3}}+…\infty\)
extreme value = \(\frac{\frac{1}{2\sqrt{3}}}{1-\frac{1}{4}}=\frac{2}{3\sqrt{3}}\)

For function f(x,y) to have no extremum value at (a,b)
rt – s² < 0 where, r = ^?2f⁄_?x², t=^?2f⁄_?y², s=^?2f⁄_?x?y, at (x,y) => (a,b).

Calculating gradients considering general form of Astroid as x^2/3 + y^2/3 = a^2/3 and n equating m by Lagrange condition.
we can conclude that maximum value increases.

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.