The problem requires us to find the modified surface area of a rectangular prism when its width and height are each shrunk to one-seventh of their original size, while the length remains unchanged.
Let's start with the original formula for the surface area of a rectangular prism, which is:
[tex]\[ \text{SA} = 2lw + 2lh + 2wh \][/tex]
Here's a step-by-step breakdown to find the modified surface area:
1. Original Dimensions:
- Length ([tex]\( l \)[/tex])
- Width ([tex]\( w \)[/tex])
- Height ([tex]\( h \)[/tex])
2. Modified Dimensions:
- Length ([tex]\( l \)[/tex]) remains the same.
- New width ([tex]\( w' \)[/tex]) = [tex]\( \frac{w}{7} \)[/tex]
- New height ([tex]\( h' \)[/tex]) = [tex]\( \frac{h}{7} \)[/tex]
3. Substitute the modified dimensions into the original surface area formula:
- Surface area formula with new dimensions:
[tex]\[ \text{SA}^{\prime} = 2l \left(\frac{w}{7}\right) + 2l \left(\frac{h}{7}\right) + 2 \left(\frac{w}{7}\right) \left(\frac{h}{7}\right) \][/tex]
4. Simplify each term:
- First term: [tex]\( 2l \left(\frac{w}{7}\right) = \frac{2lw}{7} \)[/tex]
- Second term: [tex]\( 2l \left(\frac{h}{7}\right) = \frac{2lh}{7} \)[/tex]
- Third term: [tex]\( 2 \left(\frac{w}{7}\right) \left(\frac{h}{7}\right) = 2 \left(\frac{wh}{49}\right) = \frac{2wh}{49} \)[/tex]
5. Sum up these simplified terms to get the total modified surface area:
[tex]\[ \text{SA}^{\prime} = \frac{2lw}{7} + \frac{2lh}{7} + \frac{2wh}{49} \][/tex]
So, the formula for the modified surface area is:
[tex]\[ \text{SA}^{\prime} = \frac{2lw}{7} + \frac{2lh}{7} + \frac{2wh}{49} \][/tex]
Comparing with the provided options, the correct answer is:
[tex]\[ \boxed{D. \text{SA} = \frac{2}{7} lw + \frac{2}{7} lh + \frac{2}{49} wh} \][/tex]
