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First, lets make sure we understand the word "ratio". A ratio is a comparison of any two things, usually in numerical form. We could compare the number of boys in one of my classes to the number of girls. We could compare the number of teachers to the number of students. So a ratio is simply a comparison of two things. The two things here are surface area compared to volume. Usually, the surface area will be a larger number, or represent a larger quantity than does the volume number. Lets say we have a cube that is 2 meters in length, width, and height. To get the surface area of the 6 sides of the cube, multiply the length times the width, which would be 4 square meters. then multiply times 6 for the 6 sides, which would be 24 square meters. To get the volume, multiply the length times the width times the height, which would be 8 cubic meters. So the ratio here would be 24 to 8, or 3 to 1. We would express this mathematically as 3:1.
Calculate the surface area of a rectangular prism which is 2 cm × 4 cm × 6 cm.
The surface-area-to-volume ratio is calculated by dividing the surface area by the volume of any object. If you know the formula for the surface area and the volume of an object, then simply compute (surface area) / (volume) to calculate the surface-area-to-volume ratio. The actual surface-area-to-volume ratio of any object depends upon that object's shape and geometry.
Consider a cube with equal sides of length x. The cube has six faces (top, bottom, left, right, front, back), and each face has a surface area of x2, so the total surface area of the cube is 6x2. The volume of the cube is x3. So the surface-area-to-volume ratio for a cube is 6x2 / x3, which can be reduced to 6/x. This surface-area-to-volume ratio, 6/x, holds true for all cubes.
Let's test this ratio. Consider a cube that has a 1 cm length on all sides. The surface area is 6 sides of 1 cm x 1 cm (6 cm2), and the volume is 1 cm x 1 cm x 1 cm (1 cm3). Dividing the surface area by the volume gives a surface-area-to-volume ratio of 6 (which is 6/1). If the length of the cube sides is 6 cm, then the surface area is 6 sides of 6 cm x 6 cm (216 cm2) and the volume is 6 cm x 6 cm x 6 cm (216cm3), so the surface-area-to-volume ratio is 216/216, or 1 (which is 6/6). If the length of the cube side is 12 cm, then the surface area is 6 sides of 12 cm x 12 cm (864 cm2) and the volume is 12 cm x 12 cm x 12 cm (1728 cm3), so the surface-area-to-volume ratio is 864/1728, or 0.5 (which is 6/12). We can empirically verify that this surface-area-to-volume ratio for a cube is therefore 6/x.
Consider a sphere (a round ball) of radius r. The surface area is 4 PIr2, whereas the volume is (4/3)PIr3. So the surface-area-to-volume ratio of a sphere is (4 PIr2) / [(4/3)PIr3], which can be reduced to 3/r. As in the cube, the surface-area-to-volume ratio of 3/r holds true for all spheres. In the previous description, the symbol 'PI' is meant to represent Pi, or 3.1415 ... T
For irregular objects, such as a rectangular prism (a box) with different lengths in each dimension, the surface-area-to-volume ratio must be calculated for each shape. Consider a box with dimensions of l (length), w (width), and h (height). Like the cube, the box has six faces, but it is easier to consider it as three face pairs (front/back, left/right, and top/bottom). The surface area of both faces in a pair are the same (the front face has the same surface area as the back face). So the surface area of the box is:
A = 2(l x w) + 2(w x h) + 2(l x h), or 2( (l x w) + (w x h) + (l x h) ).
The volume is:
V = l x w x h
So the surface area to volume ratio (A/V) of a box is:
2( (l x w) + (w x h) + (w x h) ) / (l x w x h).
The surface-area-to-volume ratio of a cylinder (like a soup can) is:
( (2 PI r2) + (2 PI r h) ) / (PI r2h)
Where r is the radius of the circle on the top and bottom of the cylinder, h is the height of the cylinder, and. PI is Pi, or 3.1415 ...
Unlike regular objects, such as the cube or sphere, no further simplification of the box's or cylinder's surface-area-to-volume ratio equation exists. The above appropriate equations must be applied to each box or cylinder separately.
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