Verified answer

um, the paragraph is up to you!

printed area is L by w

entire poster is L + 6 by w + 4 (since the margins will be on all sides)

Area of entire poster is (L + 6)(w + 4) = Lw + 6w + 4L + 24

and we know that Lw = 150 ==> L = 150 / w

sub to give Area = 150 + 6w + 4*150 / w + 24

Area = 6w + 600 / w + 174, the function to minimize

Area ' = 6 - 600 / w^2

set this equal to 0 ==> 6 - 600 / w^2 = 0

6 = 600 / w^2

w^2 = 100

w = 10

so the width should be 10, the the length will be 150 / 10 = 15

the smallest poster that can be used will have a printed area of 10 x 15 inches, and the entire poster will be (10 + 4) by (15 + 6) = 14 by 21

you can confirm this is a min with the second derivative test

Area" = 1200 / w^3, which is positive for positive width, confirming this is a min