optimization?

Question

an advertisement consists of a rectangular printed region plus 1-in. margins on the sides and 2-in margins at top and bottom. If the area of the printed region is to be 92 in^2, find the dimensions of the printed region and overall advertisement that minimize the total area

Answer 1

let width of printed matter be x and height be y
then its area = xy =92=> y=92/x
with margins , the area: (x+2)(y+4)
total area, f = xy+2y+4x+8
substitute y=92/x:
f = 92+8+184/x + 4x = 4x+ 184/x +100
differentiate wrt x:
df/dx = 4 - 184/x^2
for minimum value put df/dx =0

df/dx = 4 - 184/x^2 =0
=> x^2 = 184/4 = 46
x = sqrt(46)

hence y = 92/x =92/sqrt46 = 2sqrt(46)
therefore dimensions of printed area are : sqrt(46) and 2sqrt(46) units
and of overall advt: 2+sqrt(46) and 4+2sqrt(46) units respectively

Answer 2

Lel be length of printed area
let w = width of printed area
Then l*w =92 in^2

The area of the entire poster 92 + 8 +2l + 4w = 100 + 2l +4w
Substitute l = 92/w to get
A = 100 +184/w +4w
dA/dw = -184/w^2 + 4
Set this equal 0 to get -184/w^2 = -4
w^2 = -184/(-4) = 46, so w = sqrt(46)
l= 92/w = [92sqrt(46)]/46 = 2sqrt(46)
The overall dimensions including the borders is
W = sqrt(46) + 2 and L = 2sqrt(46) + 4