For a rectangular prism, if the base length was shrunk by a factor of two-thirds, the width was quadrupled, and the height was cut in half, how would the changes affect the surface area formula?

A. [tex]S A=16 h w+\frac{1}{3} l w+\frac{2}{3} w h[/tex]
B. [tex]S A=\frac{16}{3} h w-\frac{3}{4} l h+2 w h[/tex]
C. [tex]S A=\frac{16}{3} h w+\frac{2}{3} l h+4 w h[/tex]
D. [tex]S A=\frac{2}{3} h+\frac{16}{3} l w+8 w h[/tex]

Please select the best answer from the choices provided.



Answer :

Let's break down how the changes in the dimensions of a rectangular prism affect its surface area.

1. Original Surface Area Formula:
For a rectangular prism with length [tex]\( l \)[/tex], width [tex]\( w \)[/tex], and height [tex]\( h \)[/tex], the surface area [tex]\( S_A \)[/tex] is calculated as:
[tex]\[ S_A = 2lw + 2lh + 2wh \][/tex]

2. Changes in Dimensions:
- The length [tex]\( l \)[/tex] is shrunk by a factor of two-thirds, so the new length is [tex]\( \frac{2}{3}l \)[/tex].
- The width [tex]\( w \)[/tex] is quadrupled, so the new width is [tex]\( 4w \)[/tex].
- The height [tex]\( h \)[/tex] is cut in half, so the new height is [tex]\( \frac{1}{2}h \)[/tex].

3. New Surface Area Calculation:
Substituting the new dimensions into the surface area formula:
[tex]\[ S_A = 2 \left( \frac{2}{3}l \right) (4w) + 2 \left( \frac{2}{3}l \right) \left( \frac{1}{2}h \right) + 2 (4w) \left( \frac{1}{2}h \right) \][/tex]

Simplifying each term:
- For the first term:
[tex]\[ 2 \left( \frac{2}{3}l \right) (4w) = 2 \left( \frac{8}{3}lw \right) = \frac{16}{3}lw \][/tex]
- For the second term:
[tex]\[ 2 \left( \frac{2}{3}l \right) \left( \frac{1}{2}h \right) = 2 \left( \frac{2}{6}lh \right) = \left( \frac{2}{3}lh \right) \][/tex]
- For the third term:
[tex]\[ 2 (4w) \left( \frac{1}{2}h \right) = 2 (2wh) = 4wh \][/tex]

Combining all the simplified terms, the new surface area formula is:
[tex]\[ S_A = \frac{16}{3}lw + \frac{2}{3}lh + 4wh \][/tex]

4. Matching with the Given Choices:
Compare the derived formula with the given options.
The correct formula is:
[tex]\[ S_A = \frac{16}{3}lw + \frac{2}{3}lh + 4wh \][/tex]

Thus, the best answer from the provided choices is:
[tex]\[ \boxed{C. \ S A = \frac{16}{3}lw + \frac{2}{3}lh + 4wh} \][/tex]