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klulmh
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Nov 23 2011, 03:25 AM
Post #1
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Administrator
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- Nov 14, 2011
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- Quote:
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The management of a local Target store has decided to enclose an 800 square foot area outside the building for the garden display. One side will be formed by an external wall of the store, two sides will be constructed of pineboards costing $6 per foot and the side opposite the store will be constructed of fencing that costs $3 per foot. What dimensions of the enclosure will minimize the cost? Let x be the length of the side with fencing.Spoiler: click to toggle A = xy = 800 ft2
C = 2*6y + 3x = 12y + 3x
x = 800/y
C = 12y + 2400y-1
C' = 12 - 2400y-2 = 0
12 = 2400y-2
y2 = 200
y = sqrt(200) = 10sqrt(2)
when y > 10sqrt(2), C' is negative
when y < 10 sqrt(2), C' is positive
10sqrt(2) * x = 800
x = 80/sqrt(2) = 40sqrt(2)
10sqrt(2) foot pineboard by 40sqrt(2) foot fence
- Quote:
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A closed rectangular box is to be constructed with a surface area of 48 square feet so that its length is twice the width. What dimensions will maximize the volume of the box? What is the maximum volume?Spoiler: click to toggle SA = 2LW + 2LH + 2WH = 48 ft2
L = 2W
SA = 4W2 + 4WH + 2WH = 2W(2W + 2H + H) = 2W(2W +3H) = 48
3H + 2W = 24/W
3H = 24/W - 2W
H = 8/WH - 2W/3H
V = LWH = 2W*W*H = 2W2H
Edited by klulmh, Nov 23 2011, 03:40 AM.
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