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The sphere is centered at the origin and has a radius of 14 . Its equation is rewritten as x 2 + y 2 + z 2 14 = 0 . We can use the graphics primitive Sphere to construct its graph. (See 5.) f[x_, y_, z_] = x2 + y2 + z2 14; g1 = Graphics3D Sphere {0,0,0}, 14 ; a = Derivative[1, 0, 0][f][1, 2, 3]; b = Derivative[0, 1, 0][f][1, 2, 3]; c = Derivative[0, 0, 1][f][1, 2, 3]; Solve[a (x 1) + b (y 2) + c (z 3) 0, z] z 1 (14 x 2 y) 3 g2 = Plot3D 1 (14 x 2 y),{x, 5,5},{y, 5,5} ; , 3 Show[g1, g2]
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10.2 Maximum and Minimum Values
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A function, f , has a relative (or local) maximum at (x0, y0) if there exists an open disk centered at (x0, y0) such that f(x, y) f (x0, y0) for all (x, y) in the disk. A similar definition (with the inequality reversed) holds for a relative minimum. If f has either a relative maximum or relative minimum at (x0, y0), we say that f has a relative extremum at (x0, y0). If f is differentiable, a necessary condition for f(x, y) to have a relative extremum at the point (x0, y0) is fx(x0, y0) = fy(x0, y0) = 0. The point (x0, y0) is called a critical point of f.
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EXAMPLE 5 To find the critical point(s) for the function f(x, y) = x4 + y4 4xy, we compute the first-order partial
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derivatives, set them both equal to 0, and solve the resulting equations.
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Multivariate Calculus
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f[x_, y_] = x4 + y4 4 x y pdx = D[f[x, y], x] 4 x3 4 y pdy = D[f[x, y], y] 4 x + 4 y3 Solve[{pdx 0, pdy 0}, {x, y}]
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{{x 1, y 1},{x 0, y 0},{x , y }, {x , y }, {x 1, y 1},{x ( 1)1/4, y ( 1)3/4}, {x ( 1)1/4, y ( 1)3/4}, {x ( 1)3/4, y ( 1)1/4}, {x ( 1)3/4, y ( 1)1/4}}
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The only real critical points are ( 1, 1), (0, 0), and (1, 1).
Not all critical points turn out to be relative extrema. To determine whether a function has a relative extremum at a critical point, and if so, whether it is a maximum or minimum, we use the Second Partial Derivatives Test:
Let D( x , y) = fx x ( x , y) fy y ( x , y) [ fx y ( x , y)]2 and let (x0, y0) be a critical point of f.
1. If D(x0, y0) > 0 and fx x (x0, y0) > 0, then f has a relative minimum at (x0, y0). 2. If D(x0, y0) > 0 and fx x (x0, y0) < 0, then f has a relative maximum at (x0, y0). 3. If D(x0, y0) < 0, then f has neither a relative maximum nor a relative minimum at (x0, y0). We say that f has a saddle point at (x0, y0). If D(x0, y0) = 0, the test is inconclusive.
EXAMPLE 6
Continuing with the previous example, we define D(x, y). (We use d to avoid conflict with D, Mathematica s derivative operator.)
2 d[x_, y_]= {x,2} f[x, y] {y,2} f[x, y] ( x, y f[x, y]);
d[0, 0] 16 d[1, 1] 128 {x, 2}f[x, y] /.{x 1, y 1} 12 d[ 1, 1] 128 {x, 2}f[x, y] /.{x 1, y 1} 12
Relative minimum at ( 1, 1). Relative minimum at (1, 1). Negative number; saddle point at (0, 0).
It is certainly worthwhile plotting this function. Mathematica makes it easy, although some experimentation with the options is necessary to show the details clearly. Plot3D[f[x, y], {x, 2, 2}, {y, 2, 2}, PlotRange { 2, 5}, ViewPoint {1.761, 2.816, 0.647}]
2 4 2 0 2 2 1 0 1 2