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?
1 answer:
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.
You might be interested in
Answer:
-5,6
Step-by-step explanation:
cause you stared at 6 then you go down
Answer:
hi
Step-by-step explanation:
Answer:
$9.56
Step-by-step explanation:
$13.00 x .30 = 3.90
13.00 - 3.90 = $9.10 x .05 = 0.46
$9.10 + 0.46 = 9.56
Answer:
D. 17.3
Step-by-step explanation:
50=3x-2
52=3x
x=17.333
D
2x^3
2 can fit into 10 and 46
and you can take 3 out of 3 and 5
