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cestrela7 [59]
3 years ago
10

A manufacturer wants to double the volume of a 3 in.×2 in.×6 in. 3 i n . × 2 i n . × 6 i n . box, while using as little extra ca

rdboard as possible. Which statement is true?
Mathematics
1 answer:
viktelen [127]3 years ago
6 0

Answer: 2 inch dimension will give smallest increase.


Step-by-step explanation:

Length = 3 in

width = 2 in

height = 6 in

Extra cardboard  means to find surface area

on doubling the length

length = 6 In

width = 2 In

Height = 6In

Surface area for the above dimensions = 2 [ 6x2+2x6+6x6] = 120 sq in

On doubling the width

length = 3 in

width = 4 in

Height = 6 inch

Surface area for the above dimensions= 2 [ 3x4+4x6+6x3] = 2[54] = 108 sq inches

On doubling height

Length =3 in

width = 2 in

Height = 12 in

Surface area for above dimensions = 2 [ 3x2+2x12+12x3] = 2[6+24+36] = 132 sq inch

On doubling width surface area is minimum.

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There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is  14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is  14!/(23)=10,897,286,400

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