Final answer:
The maximum volume of a rectangular parallelepiped with a square base and a surface area of 384 cm^2 is found by differentiating the volume function. The maximum occurs when the base side is 8 cm, which also makes the height 8 cm, resulting in a volume of 512 cm^3, corresponding to answer C).
Step-by-step explanation:
Maximum Volume of a Rectangular Parallelepiped
To find the maximum volume of a box with a square base and given surface area, we can use differentiation to optimize the volume function.
Let the side of the square base be s and the height of the box be h. The surface area SA of the box is given by SA = 2s^2 + 4sh = 384 cm^2.
First, we solve for h in terms of s using the surface area equation:
- 2s^2 + 4sh = 384
- 4sh = 384 - 2s^2
- h = (384 - 2s^2) / (4s)
Next, we express the volume V as a function of s alone:
- V = s^2 × h
- V = s^2 × ((384 - 2s^2) / (4s))
- V = (384s - 2s^3) / 4
To find the maximum volume, we take the derivative of V with respect to s and set it equal to zero:
- dV/ds = (384 - 6s^2) / 4
- 0 = 384 - 6s^2
- s^2 = 64
- s = 8 (since s must be positive)
Substituting s = 8 cm into the equation for h, we get:
- h = (384 - 2(8)^2) / (4×8)
- h = (384 - 128) / 32
- h = 8 cm
Therefore, the maximum volume is:
- V = 8 cm × 8 cm × 8 cm = 512 cm^3
So the correct answer is C) 512 cm^3.