Answer:
[tex]6m^2+8m-16[/tex]
Step-by-step explanation:
Given that the length of the packet is [tex]l=m[/tex] cm.
Then the width being 2cm less than the length becomes [tex]w=m-2[/tex] and the height being 4 cm more than its length becomes [tex]h=m+4[/tex]
The total surface area of the rectangular packet is given by:
[tex]T.S.A=2(lw+lh+wh)[/tex]
[tex]\implies T.S.A=2[m(m-2)+m(m+4)+(m-2)(m+4)][/tex]
We expand to get:
[tex]\implies T.S.A=2(m^2-2m+m^2+4m+m^2+4m-2m-8))[/tex]
[tex]\implies T.S.A=2(3m^2+4m-8)=6m^2+8m-16[/tex]
