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 functionf(x,y)=x3−6xy+8y3

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To find the local maximum and minimum values as well as saddle points of the function f(x,y)=x3−6xy+8y3, we need to calculate the first and second partial derivatives. Let&#039;s start by finding the first partial derivatives with respect to both x and y:∂f∂x=3x2−6y∂f∂y=−6x+24y2To find the critical points, we need to solve the system of equations obtained by setting both partial derivatives equal to zero:3x2−6y=0−6x+24y2=0From the first equation, we can isolate x as follows:3x2=6yx2=2yx=±2ySubstituting this into the second equation, we get:−6(±2y)+24y2=0−62y+24y2=0Now, we can solve this equation for y. Let&#039;s multiply through by −1 to simplify it:62y−24y2=0Dividing through by 6:2y−4y2=0Squaring both sides to eliminate the square root:2y−16y4=0y(1−8y3)=0From this equation, we find two possible values for y:y=01−8y3=0The first equation gives us y=0. Substituting this back into the equation x=±2y, we find two corresponding values for x:x=±2·0=0So, one critical point is (0,0).Now, let&#039;s solve the second equation:1−8y3=0This equation gives us y=(12)13=12. Substituting this into the equation x=±2y, we find two corresponding values for x:x=±2·12=±1So, another critical point is (1,12).To determine whether these critical points are local maximum, local minimum, or saddle points, we need to calculate the second partial derivatives. Let&#039;s find them:∂2f∂x2=6x∂2f∂y2=48y∂2f∂x∂y=−6∂2f∂y∂x=−6Now, we can evaluate the second partial derivatives at the critical points we found.At (0,0):∂2f∂x2=6(&amp;lt;br&amp;gt;0)=0∂2f∂y2=48(0)=0∂2f∂x∂y=−6∂2f∂y∂x=−6The determinant of the Hessian matrix is given by:D=|0−6−60|=(0)(0)−(−6)(−6)=0−36=−36Since the determinant is negative, and the second partial derivatives test fails, we cannot make a definite conclusion about the nature of the critical point (0,0).At (1,12):∂2f∂x2=6(1)=6∂2f∂y2=48(12)=24∂2f∂x∂y=−6∂2f∂y∂x=−6The determinant of the Hessian matrix is given by:D=|6−6−624|=(6)(24)−(−6)(−6)=144−36=108Since the determinant is positive and the second partial derivatives test passes, we can conclude that (1,12) 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(x,y)=x3−6xy+8y3 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.

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

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