Question

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(x,y)=x^3-6xy+8y^3

Analyzing functions
ANSWERED
asked 2021-05-14
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.
\(f(x,y)=x^3-6xy+8y^3\)

Expert Answers (2)

2021-05-15
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Best answer
2021-09-29

With the given function, determine the critical points and determine if the critical points are local minimum, maximum or saddle point.

\(f(x,y)=x^3-6xy+8y^3\)

Identify first derivatives in terms of \(F_x(x,y)\) and \(F_y(x,y)\)

\(F_x(x,y)=3x^2-6y\)

\(F_y(x,y)=-6x+24y^2\)

Find the critical points of x by solving for y in \(F_x\) and substituting result into \(F_y\)

\(F_x(x,y)=3x^2-6y\to3x^2=6y\to\frac{1}{2}x^2=y\)

\(F_y(x,y)=-6x+24(\frac{1}{2}x^2)^2\to-6x+6x^4\)

\(=6x(x^3-1)=0\to x=0\) and \(x=1\)

Find the critical values of y associated with the critical values found for x by substituting into \(F_x\)

\(F_x(0,y)=3x^2-6y\to0=6y\to y=0\)

\(F_x(1,y)=3x^2-6y\to3=6y\to y=\frac{1}{2}\)

using \(F_x(x,y)\) and \(F_y(x,y)\) find the second derivatives in terms of \(F_{xx}(x,y),\ F_{yy}(x,y)\) and \(F_{xy}(x,y)\)

\(F_{xx}(x,y)=6x\)

\(F_{yy}(x,y)=48y\)

\(F_{xy}=-6\)

Use the second derivative test provided the rules defined.

\(D=D(x,y)=F_{xx}(x,y)\cdot F_{yy}(x,y)-(F_{xy}x,y)^2\)

\(D>0\) and \(F_{xx}(x,y)<0\), then \(F(x,y)\) is local maximum \(D<0\) then \(F(x,y)\) is a saddle point

The critical points identified as (0,0) and \((1,\frac{1}{2})\) are to be used to solve

\(D=D(0,0)=0\cdot0-(-6)^2=36;\ D=D(1,\frac{1}{2})=6\cdot24-(-6)^2=108\)

Compare D values of \(F_{xx}(x,y)\) to indentify critical points as local maximum, minimum or saddle points

\(D=D(0,0)=-36\to D<0\) Saddle point

\(D=D(1,\frac{1}{2})=108,\ F_{xx}(1,\frac{1}{2})>0\)

\(D(1,\frac{1}{2})\) is local minimum

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