To solve this problem, we need to find compatible pairs of base areas and heights that would yield the given volume \( V = 16y^4 + 16y^3 + 48y^2 \).
Using the formula for the volume of a prism, \( V = B \cdot h \), we can express this as \( B \cdot h = 16y^4 + 16y^3 + 48y^2 \).
To find possible pairs, we can use the given base areas and heights and check if their product equals the volume:
A. Base area: \( B = 4y \) square units; Height: \( h = 4y^2 + 4y + 12 \) units
\[ V = B \cdot h = 4y \cdot (4y^2 + 4y + 12) = 4y \cdot 4y^2 + 4y \cdot 4y + 4y \cdot 12 \]
Calculating each term:
- \( 4y \cdot 4y^2 = 16y^3 \)
- \( 4y \cdot 4y = 16y^2 \)
- \( 4y \cdot 12 = 48y \)
Thus:
\[ V = 16y^3 + 16y^2 + 48y \]
This is not equal to the volume \( 16y^4 + 16y^3 + 48y^2 \).
B. Base area: \( B = 8y^2 \) square units; Height: \( h = y^2 + 2y + 4 \) units
\[ V = B \cdot h = 8y^2 \cdot (y^2 + 2y + 4) = 8y^2 \cdot y^2 + 8y^2 \cdot 2y + 8y^2 \cdot 4 \]
Calculating each term:
- \( 8y^2 \cdot y^2 = 8y^4 \)
- \( 8y^2 \cdot 2y = 16y^3 \)
- \( 8y^2 \cdot 4 = 32y^2 \)
Thus:
\[ V = 8y^4 + 16y^3 + 32y^2 \]
This is not equal to the volume \( 16y^4 + 16y^3 + 48y^2 \).
C. Base area: \( B = 12y \) square units; Height: \( h = 4y^2 + 4y + 36 \) units
\[ V = B \cdot h = 12y \cdot (4y^2 + 4y + 36) = 12y \cdot 4y^2 + 12y \cdot 4y + 12y \cdot 36 \]
Calculating each term:
- \( 12y \cdot 4y^2 = 48y^3 \)
- \( 12y \cdot 4y = 48y^2 \)
- \( 12y \cdot 36 = 432y \)
Thus:
\[ V = 48y^3 + 48y^2 + 432y \]
This is not equal to the volume \( 16y^4 + 16y^3 + 48y^2 \).
D. Base area: \( B = 16y^2 \) square units; Height: \( h = y^2 + y + 3 \) units
\[ V = B \cdot h = 16y^2 \cdot (y^2 + y + 3) = 16y^2 \cdot y^2 + 16y^2 \cdot y + 16y^2 \cdot 3 \]
Calculating each term:
- \( 16y^2 \cdot y^2 = 16y^4 \)
- \( 16y^2 \cdot y = 16y^3 \)
- \( 16y^2 \cdot 3 = 48y^2 \)
Thus:
\[ V = 16y^4 + 16y^3 + 48y^2 \]
This is equal to the volume \( 16y^4 + 16y^3 + 48y^2 \).
Therefore, the correct answer is:
D. Base area: 16y^2 square units; Height: y^2 + y + 3 units.