Monverdeitum

2023-03-31

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function
$f\left(x,y\right)={x}^{3}-6xy+8{y}^{3}$

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To find the local maximum and minimum values as well as saddle points of the function $f\left(x,y\right)={x}^{3}-6xy+8{y}^{3}$, we need to calculate the first and second partial derivatives. Let's start by finding the first partial derivatives with respect to both $x$ and $y$:
$\frac{\partial f}{\partial x}=3{x}^{2}-6y$
$\frac{\partial f}{\partial y}=-6x+24{y}^{2}$
To find the critical points, we need to solve the system of equations obtained by setting both partial derivatives equal to zero:
$3{x}^{2}-6y=0$
$-6x+24{y}^{2}=0$
From the first equation, we can isolate $x$ as follows:
$3{x}^{2}=6y$
${x}^{2}=2y$
$x=±\sqrt{2y}$
Substituting this into the second equation, we get:
$-6\left(±\sqrt{2y}\right)+24{y}^{2}=0$
$-6\sqrt{2y}+24{y}^{2}=0$
Now, we can solve this equation for $y$. Let's multiply through by $-1$ to simplify it:
$6\sqrt{2y}-24{y}^{2}=0$
Dividing through by $6$:
$\sqrt{2y}-4{y}^{2}=0$
Squaring both sides to eliminate the square root:
$2y-16{y}^{4}=0$
$y\left(1-8{y}^{3}\right)=0$
From this equation, we find two possible values for $y$:
$y=0$
$1-8{y}^{3}=0$
The first equation gives us $y=0$. Substituting this back into the equation $x=±\sqrt{2y}$, we find two corresponding values for $x$:
$x=±\sqrt{2·0}=0$
So, one critical point is $\left(0,0\right)$.
Now, let's solve the second equation:
$1-8{y}^{3}=0$
This equation gives us $y={\left(\frac{1}{2}\right)}^{\frac{1}{3}}=\frac{1}{2}$. Substituting this into the equation $x=±\sqrt{2y}$, we find two corresponding values for $x$:
$x=±\sqrt{2·\frac{1}{2}}=±1$
So, another critical point is $\left(1,\frac{1}{2}\right)$.
To determine whether these critical points are local maximum, local minimum, or saddle points, we need to calculate the second partial derivatives. Let's find them:
$\frac{{\partial }^{2}f}{\partial {x}^{2}}=6x$
$\frac{{\partial }^{2}f}{\partial {y}^{2}}=48y$
$\frac{{\partial }^{2}f}{\partial x\partial y}=-6$
$\frac{{\partial }^{2}f}{\partial y\partial x}=-6$
Now, we can evaluate the second partial derivatives at the critical points we found.
At $\left(0,0\right)$:
$\frac{{\partial }^{2}f}{\partial {x}^{2}}=6\left(
0\right)=0$

$\frac{{\partial }^{2}f}{\partial {y}^{2}}=48\left(0\right)=0$
$\frac{{\partial }^{2}f}{\partial x\partial y}=-6$
$\frac{{\partial }^{2}f}{\partial y\partial x}=-6$
The determinant of the Hessian matrix is given by:
$D=|\begin{array}{cc}\hfill 0\hfill & \hfill -6\hfill \\ \hfill -6\hfill & \hfill 0\hfill \end{array}|=\left(0\right)\left(0\right)-\left(-6\right)\left(-6\right)=0-36=-36$
Since the determinant is negative, and the second partial derivatives test fails, we cannot make a definite conclusion about the nature of the critical point $\left(0,0\right)$.
At $\left(1,\frac{1}{2}\right)$:
$\frac{{\partial }^{2}f}{\partial {x}^{2}}=6\left(1\right)=6$
$\frac{{\partial }^{2}f}{\partial {y}^{2}}=48\left(\frac{1}{2}\right)=24$
$\frac{{\partial }^{2}f}{\partial x\partial y}=-6$
$\frac{{\partial }^{2}f}{\partial y\partial x}=-6$
The determinant of the Hessian matrix is given by:
$D=|\begin{array}{cc}\hfill 6\hfill & \hfill -6\hfill \\ \hfill -6\hfill & \hfill 24\hfill \end{array}|=\left(6\right)\left(24\right)-\left(-6\right)\left(-6\right)=144-36=108$
Since the determinant is positive and the second partial derivatives test passes, we can conclude that $\left(1,\frac{1}{2}\right)$ is a local minimum.
To visualize the function and confirm our results, it is recommended to graph it using three-dimensional graphing software. By plotting the function $f\left(x,y\right)={x}^{3}-6xy+8{y}^{3}$ with an appropriate domain and viewpoint, we can observe the shape of the graph and identify the critical points as well as the presence of any other important aspects such as other extrema or saddle points.

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