Find all the local maxima local minima and saddle points of

Find all the local maxima, local minima, and saddle points of the following given functions.

Type an exact answer. Type pi as needed.

a) f(x,y) = 5y sin (3x)

b) f(x,y) = x^3 + y + 9x^2 - 3y^2 - 6

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Solution

(a). we have f(x,y) = 5y sin(3x) so that f_x = 5ycos(3x) and f_y = 5 sin(3x). Thus, at the critical points , we have 5ycos(3x) = 0 …(1) and 5sin(3x) =0 or, sin(3x) = 0…(2). Now, if sin(3x) = 0, then 3x = k or, x = k/3, where k is an integer, positive, negative or zero. Also, when sin(3x) = 0 , we have cos(3x) = ±1 so that the 1^st equation changes to ± 5y = 0 or, y = 0. Thus, the critical points of the given function are (k/3,0). Further, f_xx= -5ysin(3x), f_yy= 0, f_xy= 5cos(3x) = f_yx. Then D = f_xxf_yy-f_xyf_yx = -5sin(3x)* 0- 5cos(3x)* 5cos(3x) = -25cos²(3x). Thus, D is always negative so that all the critical points (k/3,0) are saddle points. There are no local maxima or minima.

(b). we have f(x,y) = x³+y+9x²-3y²- 6 do that f_x = 3x²+18x and f_y = 1-6y. Thus, at the critical points , we have 3x²+18x = 0 or, 3x(x+6)= 0 so that either x = 0 or, x = -6   and 1-6y= 0 so that y = 1/6. Then, the critical points of the given function are (0,1/6) and (-6,1/6). Further, f_xx= 6x+18, f_yy= -6, and f_xy= 0 = f_yx. Now, D = f_xxf_yy - f_xyf_yx = (6x+18)(-6)-0 = -36(x+3). Hence, D > 0 when x < -3. Also, at the point (-6,1/6), f_xx(-6,1/6) = 3(-6)²+18(-6) = 108-108 = 0. Therefore, the point (-6,1/6) is neither a local maximum or a local minimum. Hence, the point (-6,1/6) is a saddle point. Further D < 0 when x > -3. Hence the point (0,1/6) is a saddle point. There are no local maxima or minima.