Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. The areas of the smaller rectangles are A 1 = xy, A 2 =(L − x)y,
A 3 =(L − x)(W − y), A 4 = x(W − y). For0 ≤ x ≤ L, 0 ≤ y ≤ W ,let
f(x, y) =A 2 1 + A 2 2 + A 2 3 + A 2 4
= x 2 y 2 +(L − x) 2 y 2 +(L − x) 2 (W − y) 2 + x 2 (W − y) 2
=[x 2 +(L − x) 2 ][y 2 +(W − y) 2 ]
Then we need to find the maximum and minimum values of f(x, y). Here
f x (x, y) =[2x − 2(L − x)][y 2 +(W − y) 2 ]=0 ⇒ 4x − 2L =0or x = 1 2 L,and
f y (x, y) =[x 2 +(L − x) 2 ][2y − 2(W − y)] = 0 ⇒ 4y − 2W =0or y = W/2. Also
f xx =4[y 2 +(W − y) 2 ], f yy =4[x 2 +(L − x) 2 ],andf xy =(4x − 2L)(4y − 2W ). Then
D =16[y 2 +(W − y) 2 ][x 2 +(L − x) 2 ] − (4x − 2L) 2 (4y − 2W ) 2 .Thuswhenx = 1 L and y = 1 W , D>0 and
2 2
f xx =2W 2 > 0. Thus a minimum of f occurs at 1
L, 1 W 2 2
and this minimum value is f 1
L, 1 W 2 2
= 1 4 L2 W 2 .
There are no other critical points, so the maximum must occur on the boundary. Now along the width of the rectangle let
g(y) =f(0,y)=f(L, y) =L 2 [y 2 +(W − y) 2 ], 0 ≤ y ≤ W .Theng 0 (y) =L 2 [2y − 2(W − y)] = 0 ⇔ y = 1 W . 2
And g
1
2 =
1
2 L2 W 2 . Checking the endpoints, we get g(0) = 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) 2 ], 0 ≤ x ≤ L. Bysymmetryh 0 (x) =0 ⇔ x = 1 L and 2
h 1
L = 1 2 2 L2 W 2 . At the endpoints we have h(0) = h(L) =L 2 W 2 . Therefore L 2 W 2 is the maximum value of f.
This maximum value of f occurs when the “cutting” lines correspond to sides of the rectangle.
3. (a) The area of a trapezoid is 1 2 h(b 1 + b 2 ),whereh is the height (the distance between the two parallel sides) and b 1 , b 2 are
the lengths of the bases (the parallel sides). From the figure in the text, we see that h = x sin θ, b 1 = w − 2x, and
b 2 = w − 2x +2x cos θ. Therefore the cross-sectional area of the rain gutter is
A(x, θ) = 1 2
x sin θ [(w − 2x)+(w − 2x +2x cos θ)] = (x sin θ)(w − 2x + x cos θ)
= wx sin θ − 2x 2 sin θ + x 2 sin θ cos θ, 0