Question

Find the relative maxima and relative minima values of the function $f\left( x,y \right)=8{{x}^{4}}+{{y}^{2}}-24xy$.

Answer 1

Hint: We know that for a function with multiple independent variables, we define $D={{F}_{xx}}\cdot {{F}_{yy}}-{{\left( {{F}_{xy}} \right)}^{2}}$. Now, if at critical points, $D>0$ and ${{F}_{xx}}>0$, then the point is a local minima, if $D>0$ and ${{F}_{xx}}<0$, then the point is a local maxima, if $D<0$, then the point is a saddle point and if $D=0$, we don’t have enough information to comment. Using this concept, we can find the local maxima and local minima.

Complete step-by-step answer:
We need to find the relative maxima and relative minima values for the function $f\left( x,y \right)=8{{x}^{4}}+{{y}^{2}}-24xy$.
Let us represent the differentiation of $f\left( x,y \right)$ with respect to $x$ as ${{F}_{x}}$.
So, we have
${{F}_{x}}=\dfrac{\partial }{\partial x}\left( 8{{x}^{4}}+{{y}^{2}}-24xy \right)$
Thus, we have
${{F}_{x}}=32{{x}^{3}}-24y...\left( i \right)$
Let us represent the differentiation of $f\left( x,y \right)$ with respect to $y$ as ${{F}_{y}}$.
So, we have
${{F}_{y}}=\dfrac{\partial }{\partial y}\left( 8{{x}^{4}}+{{y}^{2}}-24xy \right)$
Thus, we have
${{F}_{y}}=2y-24x...\left( ii \right)$
Now, let us equate the equation (i) with 0 to get
$32{{x}^{3}}-24y=0$
Let us simplify this equation as
$4{{x}^{3}}=3y...\left( iii \right)$
Now, let us equate the equation (ii) with 0 to get
$2y-24x=0$
Let us simplify this equation as
$y=12x...\left( iv \right)$
Let us now try to solve (iii) and equation (iv) simultaneously,
$\begin{align}
  & 4{{x}^{3}}-36x=0 \\
 & \Rightarrow x\left( {{x}^{2}}-9 \right)=0 \\
 & \Rightarrow x=0,3,-3 \\
\end{align}$
For $x=0$, we have $y=0$. And the point is (0, 0).
For $x=3$, we have $y=36$. And the point is (3, 36).
For $x=-3$, we have $y=-36$. And the point is (-3, -36).
Let us now represent the double differentiation of $f\left( x,y \right)$ with respect to $x$ as ${{F}_{xx}}$, double differentiation of $f\left( x,y \right)$ with respect to $y$ as ${{F}_{yy}}$ and differentiation of ${{F}_{x}}$ with respect to $y$ as ${{F}_{xy}}$. So, now we have
${{F}_{xx}}=\dfrac{\partial }{\partial x}\left( 32{{x}^{3}}-24y \right)$
Thus, we have
${{F}_{xx}}=96{{x}^{2}}$
${{F}_{yy}}=\dfrac{\partial }{\partial y}\left( 2y-24x \right)$
Thus, we have
${{F}_{yy}}=2$
${{F}_{xy}}=\dfrac{\partial }{\partial y}\left( 32{{x}^{3}}-24y \right)$
Thus, we have
${{F}_{xy}}=-24$
Let us define $D={{F}_{xx}}\cdot {{F}_{yy}}-{{\left( {{F}_{xy}} \right)}^{2}}$. By using the values of ${{F}_{xx}}$, ${{F}_{yy}}$ and ${{F}_{xy}}$, we get
$D=192{{x}^{2}}-576$
We know that if $D>0$ and ${{F}_{xx}}>0$, then the point is a local minima, if $D>0$ and ${{F}_{xx}}<0$, then the point is a local maxima, if $D<0$, then the point is a saddle point and if $D=0$, we don’t have enough information to comment.
So, for point (0, 0), we have
$D=192{{\left( 0 \right)}^{2}}-576=-576$
Here, since $D<0$, the point (0, 0) is a saddle point.
For point (3, 36), we have
$D=192{{\left( 3 \right)}^{2}}-576=1152$
${{F}_{xx}}=96{{\left( 3 \right)}^{2}}=864$
Here, we can see that $D>0$ and ${{F}_{xx}}>0$, so the point (3, 36) is a local minima.
For point (-3, -36), we have
$D=192{{\left( -3 \right)}^{2}}-576=1152$
${{F}_{xx}}=96{{\left( -3 \right)}^{2}}=864$
Here, we can see that $D>0$ and ${{F}_{xx}}>0$, so the point (-3, -36) is also a local minima.
Hence, local maxima does not exist, and the local minima are at points (3, 36) and (-3,-36).

Note: We must remember that when $D=0$, then this test fails and we don’t have enough information whether to tell that the point is a maxima or minima or a saddle point. We must note that for functions involving more than one independent variable, we use partial derivatives instead of normal derivatives.