Mathematics: Question is in image answer question 1

The local extremas of a function, f(x), occur when the derivative of the function, f (x), equals zero.Given the function: The derivative of the function is given by:At the local extremas, and . i.e. Recall that . . . (1) . . . (2)But from (1),Therefore, the local extremas are (0, 0) and (3, 3)...

Question is in image answer question 1 - id19053809, BreBreDoeCCx

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breannaasmith1122

27.12.2020

Mathematics

The local extremas of a function, f(x), occur when the derivative of the function, f'(x), equals zero.

Given the function: f(x, y) = 4y^3 + 18x^2 - 36xy

The derivative of the function is given by:

$12y^2f'(x)+36x-36[xf'(x)+y]=0 \\ \\ \Rightarrow12y^2f'(x)+36x-36xf'(x)-36y=0 \\ \\ \Rightarrow(12y^2-36x)f'(x)=36y-36x \\ \\ \Rightarrow f'(x)= \frac{36(y-x)}{12(y^2-3x)} = \frac{3(y-x)}{y^2-3x}$

At the local extremas, $\frac{\partial}{\partial x} =0$ and $\frac{\partial}{\partial y} =0$ . i.e. $\frac{\partial}{\partial x} =\frac{\partial}{\partial y}$

Recall that

$f'(x)= \frac{ \frac{\partial}{\partial x} }{\frac{\partial}{\partial y}} = \frac{3(y-x)}{y^2-3x} \\ \\ \Rightarrow\frac{\partial}{\partial x}=3(y-x)\ and\ \frac{\partial}{\partial y}=y^2-3x$

$\frac{\partial}{\partial x}=0 \\ \\ \Rightarrow3(y-x)=0 \\ \\ y-x=0 \\ \\ y=x$ . . . (1)

$\frac{\partial}{\partial y}=0 \\ \\ y^2-3x=0$ . . . (2)

But from (1),

$y=x \\ \\ \Rightarrow x^2-3x=0 \\ \\ \Rightarrow x(x-3)=0 \\ \\ \Rightarrow x=0\ and\ x-3=0 \\ \\ \Rightarrow x=0\ and\ x=3$

Therefore, the local extremas are (0, 0) and (3, 3)

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Question is in image answer question 1

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