Thomas Calculus 13th [Solutions]

Section 4.6 Applied Optimization 291
2
6. The area of the triangle is A
1
ba b 400 b ,
2 2
2
where 0 b 20. Then dA 1 2
400 b b
db 2
2
2 400 b
2
200 b 0 the interior critical point is b 10 2.
2
400 b
When b 0 or 20, the area is zero A 10 2 is the
2 2
maximum area. When a b 400 and b 10 2,
the value of a is also 10 2 the maximum area
occurs when a b.
7. The area is A( x) x(800 2 x ), where 0 x 400.
Solving A ( x) 800 4x 0 x 200. With
A(0) A (400) 0, the maximum area is
2
A (200) 80,000 m . The dimensions are 200 m by
400 m.
8. The area is 2xy 216 y
108
. The amount of
x
1
fence needed is P 4x 3y 4x 324 x , where
0 x;
dP 324 2
4 0 x 81 0 the critical
dx
2
x
points are 0 and 9, but 0 and 9 are not in the
domain. Then P (9) 0 at x 9 there is a
minimum the dimensions of the outer rectangle are
18 m by 12 m 72 meters of fence will be needed.
9. (a) We minimize the weight tS where S is the surface area, and t is the thickness of the steel walls of the
2
tank. The surface area is S x 4xy where x is the length of a side of the square base of the tank, and y
3
is its depth. The volume of the tank must be 500 ft y
500
. Therefore, the weight of the tank is
2
x
2
w( x) t x
2000
. Treating the thickness as a constant gives w ( x) t 2 x
2000
. The critical value is
x
2
x
4000
10
at x 10. Since w (10) t 2 0, there is a minimum at x 10. Therefore, the optimum dimensions
3
of the tank are 10 ft on the base edges and 5 ft deep.
(b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of
the steel walls would likely be determined by other considerations such as structural requirements.
3
2 1125
10. (a) The volume of the tank being 1125 ft , we have that yx 1125 y . The cost of building the
2
x
2
tank is c( x) 5x 30 x
1125
, where 0 x . Then c ( x) 10x 33750
0 the critical points are 0 and 15,
2
2
x
x
but 0 is not in the domain. Thus, c (15) 0 at x 15 we have a minimum. The values of x 15 ft and
y 5 ft will minimize the cost.
2
(b) The cost function c 5( x 4 xy) 10 xy , can be separated into two items: (1) the cost of the materials and
labor to fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of
2
2
the tanks is ( x 4 xy ), it can be deduced that the unit cost to fabricate the tanks is $5/ft . Normally,
excavation costs are per unit volume of excavated material. Consequently, the total excavation cost can be
2
2
taken as 10
10
$10/ft
xy ( x y ). This suggests that the unit cost of excavation is where x is the length of
x
x
a side of the square base of the tank in feet. For the least expensive tank, the unit cost for the excavation is
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