Find the local maximum and minimum values and saddle point(s) of the function. F

Question

Find the local maximum and minimum values and saddle point(s) of the function.
Find the local maximum and minimum values and saddle point(s) 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. (Enter NONE in any unused answer blanks.)

f(x, y) = 2x^3 + xy^2 + 5x^2 + y^2

maximum
f( ? , ? ) = ? (smaller x value)
f( ? , ? ) = ? (larger x value)
minimum
f( ? , ? ) = ? (smaller x value)
f( ? , ? ) = ? (larger x value)
saddle points
( ? , ? ) (smallest x value)
( ? , ? ) (largest x value)

Explanation / Answer

The critical points are when df/dx = 0 and df/dy = 0 ; (i) df/dx = 6x² + y² + 10x = 0 (ii) df/dy = 2xy + 2y = 0 ? 2y( x + 1) = 0 with solutions x = - 1 or y = 0 ; Substitute in (i) : x = -1 ? 6 + y² - 10 = 0 ? y² = 4 ? y = ± 2 ; y = 0 ? 6x² + 10x = 0 ? x = 0 or x = - 5/3 : So critical points are (-1,-2) ; (-1,2) ; (0 ,0) ; ( -5/3, 0) For determining the nature of the critical points ( x0 , y0 ), evaluate : D(x0,y0) = d²f/dx² * d²f/dy² - ( d²f/dxdy )² in the critical points , this way : if D > 0 and d²f/dx² in (x0,y0) > 0 , then f(x0,y0) is a relative minimum value ; if D > 0 and d²f/dx² in (x0,y0) < 0 , then f(x0,y0) is a relative maximum value ; if D < 0 , then f(x0,y0) is a saddle point. d²f/dx² = 12x + 10 d²f/dy² = 2x + 2 d²f/dxdy = 2y I see that max and min points are at (0 ,0) ; ( -5/3, 0) ; and saddle points at (-1,-2) ; (-1,2)

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