Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

D PROBLEMS PLUS
1. The areas of the smaller rectangles are A1 = xy, A2 = (L - x)y,
A 3 = (L- x)(W -y), A 4 = x(W - y). ForO :::;: x :::;: L,O :::;: y:::;: W, let
f(x, y) = Ai +A~+ A~ +A~
= x 2 y 2 + (L - x?y 2 + (L- x) 2 (W - v? + x 2 (W- y) 2
y
W-y
X
L
L-x
T
w
1
= [x2 + (L- x?JIY2 + (W - y)2]
Then we need to find the maximum and minimum values of f(x , y). Here
f,(x, y) = [2x - 2(L- x)Jiy 2 + (W - y) 2 ] = 0 => 4x '-- 2L = 0 or x = ~L, and
· j 11
(x, y) = [x 2 + (L - x) 2 JI2y- 2(W- y)] = 0 => 4y- 2W = 0 or y = W / 2. Also
f :tx = 4[y 2 + (W - u?J, f v 11 = 4(x 2 + (L- x?J, and fxv = (4x- 2L)(4y- 2W). Then.
D = 16(y 2 + (W - y) 2 J!x 2 + (L- x) 2 )- (4x- 2L) 2 (4y- 2W) 2 . Thus when x =~Landy= ~W, D > 0 and
f x:r: = 2W 2 > 0. Thus a minimum off occurs at (~L , %W) and this ·minimum value is f(~L, ~W) = ~L 2 W 2 .
There are no other critical points, so the maximum must occur on the boundary. Now along the width ofthe rectangle let
g(y) = j(O, y) = f(L, y) = L 2 [y 2 + (W- Y?]. 0 :::;: y:::;: W . Then g'(y) = L 2 [2y - 2(W- y)) = 0 ~ y = ~W .
And g(%) = ~L 2 W 2 .
Checking the endpoints, we get g(O) = g(W) = L 2 W 2 • Along the length of the rectangle let
h (x) = f(x, 0) = f(x, W) = W 2 [x 2 + (L- x?]. 0 :::;: x :::;: L. By symmetry h'(x) = 0 ~ x = ~Land
h( ~L) = %L 2 W 2 • At the endpoints we have.h(O) = h(L) = L 2 W 2 • Therefore L 2 W 2 is the maximum value of f .
This maximum value off occurs when the "cutting" lines correspond to sides of the rectangle.
3. (a) The area of a trapezoid is ~ h(b1 + b2), where his the height (the distance between the two parallel sides) and b 1 , ~ are
the Ieng!hs of the bases (the parallel sides). From the figure in the text, we see that h = x sinO, b 1 = w - 2x, and
b 2 = w - 2x + 2x cos 8. Therefore the cross-sectional area of the rain gutter is
~(x, 8) = ~xsin 0 [(w - 2x) + (w- 2x + 2xcos 8)) = (x sinO)(w - 2x + xcos8)
= wx sine - 2x 2 sin B + x 2 sin B cos e. 0 < X ::; ~w. 0 < B :::;: i
We look for the critical points of A : oAf ax = w sin B -
4x sin B + 2x sin B cos B and
EJAfEJO = wxcos B- 2x 2 cosB + x 2 (cos 2 8 - sin 2 8), so {)Ajax= 0 ~ sinO (w - 4x + 2x cosO) = 0 ~
cos B = 4 x - w = 2 -
2x
w (0 < 8 :::;: ~ => sin 8 > 0). If, in addition, oAf EJB = 0, then
2x
0 = wxcosB - 2x 2 cosO+ x 2 (2cos 2 8 - 1)
= wx ( 2.- ~) - 2x 2 ( 2 - ~) + x 2 [ 2 ( 2 - ~) 2 - 1]
= 2wx - ~w 2 - 4x 2 + wx + x 2 [ 8 - 4 : + ~: - 1] = -wx + 3x 2 = x(3x - w)
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