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Solução_Calculo_Stewart_6e

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F.<br />

TX.10<br />

SECTION 4.7 OPTIMIZATION PROBLEMS ¤ 195<br />

11. xy =1.5 × 10 6 ,soy =1.5 × 10 6 /x. Minimize the amount of fencing, which is<br />

3x +2y =3x +2(1.5 × 10 6 /x) =3x +3× 10 6 /x = F (x).<br />

F 0 (x) =3− 3 × 10 6 /x 2 =3(x 2 − 10 6 )/x 2 . The critical number is x =10 3 and<br />

F 0 (x) < 0 for 0 10 3 , so the absolute minimum<br />

occurs when x =10 3 and y =1.5 × 10 3 .<br />

The field should be 1000 feet by 1500 feet with the middle fence parallel to the short side of the field.<br />

13. Let b be the length of the base of the box and h the height. The surface area is 1200 = b 2 +4hb ⇒ h = (1200 − b 2 )/(4b).<br />

The volume is V = b 2 h = b 2 (1200 − b 2 )/4b =300b − b 3 /4 ⇒ V 0 (b) =300− 3 4 b2 .<br />

V 0 (b) =0 ⇒ 300 = 3 4 b2 ⇒ b 2 =400 ⇒ b = √ 400 = 20. SinceV 0 (b) > 0 for 0 3 45<br />

. The minimum cost is C 3 45<br />

=32(2.8125) 2/3 +180 √ 2.8125 ≈ $191.28.<br />

16 16<br />

17. The distance from a point (x, y) on the line y =4x +7to the origin is (x − 0) 2 +(y − 0) 2 = x 2 + y 2 .However,itis<br />

easier to work with the square of the distance; that is, D(x) =<br />

x2 + y 2 2<br />

= x 2 + y 2 = x 2 +(4x +7) 2 . Because the<br />

distance is positive, its minimum value will occur at the same point as the minimum value of D.<br />

D 0 (x) =2x +2(4x + 7)(4) = 34x +56,soD 0 (x) =0 ⇔ x = − 28<br />

17 .<br />

D 00 (x) =34> 0,soD is concave upward for all x. Thus,D has an absolute minimum at x = − 28<br />

17<br />

origin is (x, y) = − 28 , 4 − 28<br />

17 17<br />

<br />

+7<br />

<br />

=<br />

<br />

−<br />

28<br />

17 , 7 17<br />

<br />

.<br />

. The point closest to the<br />

19. From the figure, we see that there are two points that are farthest away from<br />

S <br />

− 1 3 =<br />

16<br />

,andS(1) = 0, we see that the maximum distance is 3<br />

<br />

y = ± 4 − 4 <br />

− 1 2<br />

= ±<br />

3<br />

A(1, 0). The distance d from A to an arbitrary point P (x, y) on the ellipse is<br />

d = (x − 1) 2 +(y − 0) 2 and the square of the distance is<br />

S = d 2 = x 2 − 2x +1+y 2 = x 2 − 2x +1+(4− 4x 2 )=−3x 2 − 2x +5.<br />

S 0 = −6x − 2 and S 0 =0 ⇒ x = − 1 3 .NowS00 = −6 < 0, so we know<br />

that S has a maximum at x = − 1 .Since−1 ≤ x ≤ 1, S(−1) = 4,<br />

3<br />

32<br />

= ± √ 4<br />

9 3 2 ≈ ±1.89. The points are −<br />

1<br />

, ± √ 4<br />

3 3 2 .<br />

16<br />

3<br />

. The corresponding y-values are

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