Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.7 OPTIMIZATION PROBLEMS ¤ 197
29. Thecylinderhassurfacearea
2(area of the base) + (lateral surface area) = 2π(radius) 2 +2π(radius)(height)
=2πy 2 +2πy(2x)
Now x 2 + y 2 = r 2 ⇒ y 2 = r 2 − x 2 ⇒ y = √ r 2 − x 2 , so the surface area is
S(x)=2π(r 2 − x 2 )+4πx √ r 2 − x 2 , 0 ≤ x ≤ r
=2πr 2 − 2πx 2 +4π x √ r 2 − x 2
Thus, S 0 (x) =0− 4πx +4π x · 1
2 (r2 − x 2 ) −1/2 (−2x)+(r 2 − x 2 ) 1/2 · 1
x 2
=4π −x − √
r2 − x + √
r 2 − x 2 =4π · −x √ r 2 − x 2 − x 2 + r 2 − x 2
√ 2 r2 − x 2
S 0 (x) =0 ⇒ x √ r 2 − x 2 = r 2 − 2x 2 () ⇒ x √ r 2 − x 2 2
=(r 2 − 2x 2 ) 2 ⇒
x 2 (r 2 − x 2 )=r 4 − 4r 2 x 2 +4x 4 ⇒ r 2 x 2 − x 4 = r 4 − 4r 2 x 2 +4x 4 ⇒ 5x 4 − 5r 2 x 2 + r 4 =0.
This is a quadratic equation in x 2 . By the quadratic formula, x 2 = 5 ± √ 5
r 2 , but we reject the root with the + sign since it
10
5 −
doesn’t satisfy (). [The right side is negative and the left side is positive.] So x =
√ 5
r. SinceS(0) = S(r) =0,the
10
maximum surface area occurs at the critical number and x 2 = 5 − √ 5
10
r 2 ⇒ y 2 = r 2 − 5 − √ 5
10
r 2 = 5+√ 5
10
r 2 ⇒
the surface area is
2π
5+ √ 5
10
r 2 +4π
5 − √
5 5+ √ √ √
5
r 2 = πr 2 2 · 5+√ 5 (5− 5)(5+ 5)
+4
= πr 2 5+ √
5
+ 2√ 20
10 10 10 10
5 5
= πr 2 5+ √ 5+2·2 √
5
= πr 2 5+5 √
5
= πr 2 1+ √ 5 .
5
5
31. xy = 384 ⇒ y =384/x. Total area is
A(x) =(8+x)(12 + 384/x) = 12(40 + x +256/x),so
A 0 (x) = 12(1 − 256/x 2 )=0 ⇒ x =16. There is an absolute minimum
when x =16since A 0 (x) < 0 for 0 16.
When x =16, y =384/16 = 24, so the dimensions are 24 cm and 36 cm.
33. Let x be the length of the wire used for the square. The total area is
x
√
2 1 10 − x 3 10 − x
A(x)= +
4 2 3 2 3
= 1 16 x2 + √ 3
36 (10 − x)2 , 0 ≤ x ≤ 10
A 0 (x) = 1 x − √ √3
3
9
(10 − x) =0 ⇔ x + 4√ 3
x − 40√ 3
=0 ⇔ x = 40√ 3
8 18 72 72 72 9+4 √ .NowA(0) = 3 36
A(10) = 100 =6.25 and A 40 √
3
≈ 2.72,so
16
9+4 √ 3
(a ) The maximum area occurs when x =10m, and all the wire is used for the square.
(b) The minimum area occurs when x = 40√ 3
9+4 √ ≈ 4.35 m.
3
100 ≈ 4.81,