If 800 feet of fencing is used to enclose a rectangular plot of land that borders a river, what is the maximum area that can be
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We have no dimensions to work with. I'll pick some and try and comply with the conditions of the problem.
Suppose you have an object that is 14 by 22 by 27 cm. These three numbers have no common factor so they cannot be reduced any further, which is helpful for this problem.
Find the Volume
Volume
l = 27 cm
w = 14 cm
h = 22 cm
V = 27 *14 * 22
V = 8316 cm^3
Find the surface area
SA = 2*l*w + 2*l*h + 2*w*h
SA = 2*27*14 + 2*27*22 + 2*14*22
SA = 756 + 1188 + 616
SA = 2558
Just looking at these numbers The surface area is about 1/3 of the volume. I don't think this is always true.
Another way to do this is to consider a cube which might give you a more useful result.
s = L = W = H all three dimensions are equal in a cube.
The volume of a cube is s*s*s = s^3
The surface area of a cube is 2*s*s + 2*s*s + 2s*s = 6s^2
That means whatever the side length, the Surface Area to volume = 6/the side length which is kind of an interesting result.
Suppose you have an object that is 14 by 22 by 27 cm. These three numbers have no common factor so they cannot be reduced any further, which is helpful for this problem.
Find the Volume
Volume
l = 27 cm
w = 14 cm
h = 22 cm
V = 27 *14 * 22
V = 8316 cm^3
Find the surface area
SA = 2*l*w + 2*l*h + 2*w*h
SA = 2*27*14 + 2*27*22 + 2*14*22
SA = 756 + 1188 + 616
SA = 2558
Just looking at these numbers The surface area is about 1/3 of the volume. I don't think this is always true.
Another way to do this is to consider a cube which might give you a more useful result.
s = L = W = H all three dimensions are equal in a cube.
The volume of a cube is s*s*s = s^3
The surface area of a cube is 2*s*s + 2*s*s + 2s*s = 6s^2
That means whatever the side length, the Surface Area to volume = 6/the side length which is kind of an interesting result.