The local extremas of a function, f(x), occur when the derivative of the function, f'(x), equals zero.
Given the function: [tex]f(x, y) = 4y^3 + 18x^2 - 36xy[/tex]
The derivative of the function is given by:
[tex]12y^2f'(x)+36x-36[xf'(x)+y]=0 \\ \\ \Rightarrow12y^2f'(x)+36x-36xf'(x)-36y=0 \\ \\ \Rightarrow(12y^2-36x)f'(x)=36y-36x \\ \\ \Rightarrow f'(x)= \frac{36(y-x)}{12(y^2-3x)} = \frac{3(y-x)}{y^2-3x} [/tex]
At the local extremas, [tex] \frac{\partial}{\partial x} =0[/tex] and [tex] \frac{\partial}{\partial y} =0[/tex]. i.e. [tex] \frac{\partial}{\partial x} =\frac{\partial}{\partial y} [/tex]
Recall that
[tex]f'(x)= \frac{ \frac{\partial}{\partial x} }{\frac{\partial}{\partial y}} = \frac{3(y-x)}{y^2-3x} \\ \\ \Rightarrow\frac{\partial}{\partial x}=3(y-x)\ and\ \frac{\partial}{\partial y}=y^2-3x[/tex]
[tex]\frac{\partial}{\partial x}=0 \\ \\ \Rightarrow3(y-x)=0 \\ \\ y-x=0 \\ \\ y=x[/tex] . . . (1)
[tex]\frac{\partial}{\partial y}=0 \\ \\ y^2-3x=0[/tex] . . . (2)
But from (1),
[tex]y=x \\ \\ \Rightarrow x^2-3x=0 \\ \\ \Rightarrow x(x-3)=0 \\ \\ \Rightarrow x=0\ and\ x-3=0 \\ \\ \Rightarrow x=0\ and\ x=3[/tex]
Therefore, the local extremas are (0, 0) and (3, 3)
