Answer:
The correct option is B: the surface area of the second box is 4 times larger.
Step-by-step explanation:
The surface area of the first box is given by:
[tex] SA_{1} = 2l_{1}h_{1} + 2w_{1}h_{1} + 2l_{1}w_{1} [/tex]
Where:
l₁: is the length = 8 in
h₁: is the height = 15 in
w₁: is the width = 6 in
[tex] SA_{1} = 2*8*15 + 2*6*15 + 2*8*6 = 516 in^{2} [/tex]
The second box's dimensions are:
l₂ = 2*8 in = 16 in
h₂ = 2*15 in = 30 in
w₂ = 2*6 in = 12 in
Hence, the surface area of the second box is:
[tex]SA_{2} = 2*2l_{1}*2h_{1} + 2*2w_{1}*2h_{1} + 2*2l_{1}*2w_{1} = 4(2l_{1}h_{1} + 2w_{1}h_{1} + 2l_{1}w_{1}) = 4*SA_{1} = 4*516 in^{2} = 2064 in^{2}[/tex]
Therefore, the correct option is B: the surface area of the second box is 4 times larger.
I hope it helps you!
