To find the maximum volume of a pyramid that can fit inside a cube, we need to consider the dimensions of the pyramid that would give the largest volume.
Let's assume that the pyramid is a square pyramid with a square base of side length x. Since the pyramid has to fit inside a cube with a side length of 12cm, we know that x ≤ 12.
The height of the pyramid, h, can be found using the Pythagorean theorem applied to the right triangle formed by the height, half the side of the base, and a side edge of the pyramid. We have:
h^2 + (x/2)^2 = x^2
h^2 = x^2 - (x/2)^2
h^2 = 3x^2/4
h = √(3x^2/4)
The volume of the pyramid is given by:
V = (1/3) * x^2 * h
Substituting h:
V = (1/3) * x^2 * √(3x^2/4)
Simplifying:
V = √3 x^3 / 12
To find the maximum volume, we need to find the value of x that gives the maximum value of V. Since x ≤ 12, we can take the derivative of V with respect to x and set it to zero:
dV/dx = (√3/4) * x^2 - (√3/8) * x^2 = (√3/8) * x^2
Setting dV/dx = 0:
(√3/8) * x^2 = 0
x = 0 or x ≠ 0
Since x has to be greater than 0, we have x ≠ 0. Therefore, the only critical point is at x = 0.
Now we need to check the endpoints of the interval [0, 12]. We have:
V(0) = 0
V(12) = √3 * 12^3 / 12 = 144√3
Therefore, the maximum volume of a square pyramid that can fit inside a cube with a side length of 12cm is 144√3 cubic centimeters.
What is the maximum volume of a pyramid that can fit inside a cube with a side length of 12 cm?
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