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
