Answer:
L  =  23 in
h  = 46  in
Step-by-step explanation:
Let call  " x "  and  " y "  dimensions of the print area of the poster then:
882 = x*y  and   y = 882/x
We also know that dimensions of the poster is:
L  =  x + 2   in  and     h  =  y + 4 in
Therefore area of the poster is:
A(p)  = ( x + 2 ) * ( y + 4 )  And area as function of x is:
A(x)  =   ( x + 2 ) * ( 882/x + 4 )
A(x)  =  882 + 4*x + 1764 /x  + 8
Taking derivatives on both sides of the equation we have:
A´(x)  =  4  - 1764/x²
A´(x)  = 0      ⇒      4  - 1764/x² = 0      ⇒ 4*x² - 1764  =  0
x²  =  1764 / 4
x²  = 441 
x  = 21 in      and    y  =  882/x         y  = 42
The second derivative  A´´(x)  is > 0  ( the second term changes its sign to +) there is a minimum for the function at the point      x = 21
As x   and  y are dimensions of the printing area of the poster, dimensions of the poster  are
L  = x  +  2   =   21  + 2  =  23 in   and
h = y  +  4   =   42  + 4  =  46  in