Download Math 131. Applied Optimization Problems Name

Math 131. Applied Optimization Problems
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1. (a) An open box of maximum volume is to be made from a square piece of cardboard, 12
inches on side, by cutting squares from each corner and turning up the sides (see figure). Find
the dimensions of the box of maximum volume.
(b) Suppose the square piece of cardboard has a side of w units. What will the dimensions of
the box of maximum volume be? Your answer should be in terms of w?
Answer. (a) Let x be the side of each corner that is cut out. Then the height of the open box
will be x, and the base and width are each 12 − 2x. Then
V = x(12 − 2x)(12 − 2x) = 144x − 48x2 + 4x3
Taking the derivative, we find that
V 0 (x) = 144 − 96x + 12x2 = 12(12 − 8x + x2 ) = 12(x − 6)(x − 2).
The critical number x = 6 yields V = 0. The critical number x = 2 leads to the dimensions of
maximum volume: 8 inches by 8 inches by 2 inches.
(b) When the cardboard has side w units where w is a constant, we have
V = x(w − 2x)(w − 2x) = w2 x − 4wx2 + 4x3 .
Then V 0 (x) = w2 − 8wx + 12x2 = (w − 2x)(w − 6x). This leads to critical numbers x = w/2
and x = w/6. The critical number w/2 yields V = 0. The critical number x = w/6 leads to
the dimensions of maximum volume:
2w
w
2w
units by
units by
units.
3
3
6
2. Find two positive numbers such the sum of the first number and twice the second number
is 216 and the product is a maximum.
Answer. Let the first number be x and the second number be y. Then x + 2y = 216. Thus
x = 216 − 2y and the product function is
P (y) = xy = (216 − 2y)y = 216y − 2y 2 .
Then P 0 (y) = 216 − 4y and so the critical number is y = 216/4 = 54. The maximum product
is then obtained when the first number is 108 and the second number is 54.
3. (a) A rectangular page is to contain 30 square inches of print. The margins are one inch on
all edges. Find the dimensions of the page such that the least amount of paper is used.
(b) Suppose the constraints in (a) are changed so that top margin will be 2 inches and the side
and bottom margins will remain at one inch. Find the dimensions of the page such that least
amount of paper is used.
Answer. (a) Let the width of the print on the page be x, and the height be y. Then xy = 30
and so y = 30/x. The area of the page is
A(x) = (x + 2)(30/x + 2) = 30 + 60x−1 + 2x + 4, x > 0
√
Then A0 (x) = −60x−2 + 2 and solving −60x−2 +√2 = 0 or x2 = 30 and x = 30 is√the critical
√
30/ 30 = 30number of A(x). Then the width of the print is 30-inches and the height is √
inches.
√ Then the dimension of the page using the least amount of paper are ( 30 + 2)-inches
by ( 30 + 2)-inches
−1
0
−2
(b) In this case A(x) = (x + 2)(30/x + 3)
√ = 36 + 60x + 3x and so A (x)
√ = −60x + 3.
The √
relevant critical number is then x = 20, and we obtain dimensions ( 20 + 2)-inches by
(30/ 20 + 2)-inches
4. Find the point(s) on the parabola y = 9 − x2 that are closest to the point (0, 7).
Answer. The distance from a point (x, y) = (x, 9 − x2 ) on the parabola to (0, 7) is given by
p
√
√
d(x) = (x − 0)2 + (9 − x2 − 7)2 = x2 + 4 − 4x2 + x4 = x4 − 3x2 + 4.
The minimum of d(x) occurs when f (x) = x4 − 3x2 + 4 has a minimum. Then
f 0 (x) = 4x3 − 6x = 2x(2x2 − 3)
r
3
. There is a relative maximum when x = 0.
The critical numbers of f are x = 0 and x = ±
2
r
3
The function f (x) has minimums at the numbers x = ±
. Thus the points on the parabola
!
!2
r
r
3 15
3 15
that are closest to (0, 7) are
,
and −
,
2 2
2 2
5. Find the rectangle of maximum area that can be inscribed in a semi circle of radius r > 0.
Answer. The base of √
the rectangle as in the figure is 2x and the height is y =
the area is A(x) = 2x r2 − x2 . Then using the product and chain rules
√
r2 − x2 . Thus
A0 (x) = 2(r2 − x2 )1/2 − 2x2 (r2 − x2 )−1/2 = 2(r2 − x2 )−1/2 (r2 − 2x2 )
√
Thus the critical number is found by√solving r2 −2x2 = 0 and so√x = r/ 2. Then the rectangle
with maximum area has a base of 2r units and a height of 2r/2 units (and an area of r2
square units).
6.
A rectangular package that can be sent by the postal service can have a maximum
combined length and girth (perimeter of a cross section) of 108 inches. Find the dimensions of
the package of maximum volume that can be sent (assume the cross section is square).
Answer. With reference to the figure, we deduce y + 4x = 108, and the volume of the package
is V = x2 y. The equation y + 4x = 108 implies y = 108 − 4x. Then the volume is the function
V (x) = x2 (108 − 4x) = 108x2 − 4x3 .
Then V 0 (x) = 216x − 12x2 = 12x(18 − x). The critical numbers are x = 0 and x = 18. The
value x = 0, leads to V = 0, and the value x = 18 leads to the dimensions of maximum volume:
18 inches by 18 inches by 36 inches.