Thomas Calculus 13th [Solutions]

1040 Chapter 14 Partial Derivatives
fx
fy
D[f[x,y], x];
D[f[x,y], y];
critical Solve[{fx 0, fy 0},{x, y}]
fxx
fxy
fyy
D[fx, x];
D[fx, y];
D[fy, y];
2
discriminant fxx fyy fxy
{{x, y}, f[x, y], discriminant, fxx} /.critical
14.8 LAGRANGE MULTIPLIERS
1. f yi xj and g 2xi 4yj so that f g yi xj (2xi 4 yj ) y 2x
and x 4y
2 2
x 8x or x 0.
4
CASE 1: If x 0, then y 0. But (0, 0) is not on the ellipse so x 0.
2
2
2
CASE 2: x 0 x 2y 2y 2y 1 y 1 .
4 2
2
Therefore f takes on its extreme values at 1
2
, and , 1 . The extreme values of f on the ellipse
2 2
2 2
2
are
2 .
2. f yi xj and g 2xi 2yj so that f g yi xj (2xi 2 yj ) y 2x
and x 2y
x 4x 2
x 0 or 1
2 .
CASE 1: If x 0, then y
2
0. But (0, 0) is not on the circle x
2
y 10 0 so x 0.
2 2
CASE 2: x 0 1 y 2x 1 x x x
2 2
10 0 x 5 y 5.
Therefore f takes on its extreme values at 5, 5 and 5, 5 . The extreme values of f on the circle
are 5 and 5.
3. f 2xi 2yj and g i 3j so that
3 3 10 2 x 1 and y 3
2 2
extreme value is f (1, 3) 49 1 9 39.
f g 2xi 2 yj ( i 3 j ) x and y 3
2
2
f takes on its extreme value at (1, 3) on the line. The
4.
2
2
f 2xyi x j and g i j so that f g 2 xyi x j ( i j ) 2xy
and
2
2xy x x 0 or 2 y x.
CASE 1: If x 0, then x y 3 y 3.
CASE 2: If x 0, then 2y x so that x y 3 2y y 3 y 1 x 2.
Therefore f takes on its extreme values at (0, 3) and (2, 1). The extreme values of f are f (0, 3) 0 and
f (2, 1) 4.
2
x
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