Answer:
Poster dimensions:
w = 18 in
h = 36 in
A(min) = 648 in²
Explanation:
Printed area for the rectangular poster is:
512 in²
If we call "x" and "y" dimensions of printed area we have:
512 = x*y ⇒ y =512/x
According to problem statment dimensions of the poster will be:
w = x + 2 and h = y + 4
Then area of the poster is:
A(p) = w*h ⇒ A(p) = ( x + 2 ) * ( y + 4 )
A(p) = x*y + 4*x + 2*y + 8
And as y = 512/x we can express A(p) as a function of x
A(x) = x* (512/x) + 4*x + 2*(512/x) + 8
A(x) = 512 +4*x + 1024/x + 8 ⇒ A(x) = 520 +4*x + 1024/x (1)
Taking derivatives on both sides of the equation (1) we get:
A´(x) = 4 - 1024/x²
A´(x) = 0 ⇒ 4 - 1024/x² = 0
4*x² = 1024
x² = 1024/4
x² = 256
x = 16 in and y = 512/16 y = 32 in
The value of A´(16) = 0 since 1024 /256 = 4
And A´´(x) = - [ - 1024(2x)/ x⁴] will be A´´(x) > 0
Then the function has a minimum for x = 16
And dimensions of the poster are:
w = x + 2 w = 18 in
h = y + 4 y = 36 in
A (min) = 36*18
A (min) = 648 in²