Minimum, Maximum Value of a Two-variable Function
Date: 6/27/96 at 21:43:5 From: Anonymous Subject: Minimum, Maximum Value of Two-variable Function Find f min and f max where x and y are real numbers and f(x,y) = 2(sin x)(cos y)+3(sin x)(siny)+6(cos x). Thanks, Michael
Date: 6/28/96 at 14:6:1 From: Doctor Jerry Subject: Re: Minimum, Maximum Value of Two-variable Function Because of periodicity, it is sufficient to look at the square [0,2pi]x[0,2pi]. It is easy to check that on the boundaries, -2sqrt(10)<=f(x,y)<=2sqrt(10). For example, on the y=0 and y=2pi boundaries, f[x,0] = f[x,2pi] = 2sin x +6 cos x = 2sqrt(10)sin(x+a). On the x = 0 and x = 2pi boundaries, f[0,y] = f[2pi,y] = 6. Setting the two partial derivatives equal to zero gives 2cos x cos y-6 sin x +3 cos x sin y = 0 3cos y sin x-2sin x sin y = 0. To solve the second equation, we want to assume that sin x is not zero. Note that on the lines where sin x = 0, f = 6. If sin x is not zero, then from second equation, y = arctan(3/2) or pi+arctan(3/2). From first equation, x = arctan(sqrt(13)/6) or x = pi+arctan(sqrt(13)/6). Evaluating f at these four points and recalling the boundary values, it follows that max f is 7 and min f is -7. Moreover, f(arctan(sqrt(13)/6),arctan(3/2)) = 7 f[pi+arctan(sqrt(13)/6),arctan(3/2)) = -7 -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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