VERY DIFFICULT CALCULUS QUESTION

You are designing a rectangular poster to contain 60 sq. in. of printing with a 4 in. margin at the top and bottom and a 1 in. margin at each side. What overall dimensions will minimize the amount of paper used?

Base of the entire poster_______in (not the print area)
Height of the entire poster_______in (not the print area)
Area of the entire poster_______in (not the print area)






thanks

If x is the base of the poster and y is its height and the base with margin is x+2 and the height with margin is y+8.You have

A=(x+2)(y+8)

But xy=60 then y=60/x

A=(x+2)(60/x +8)=(x+2)((60+8x)/x)

A=60+ 8x+120/x+ 16=8x+120/x+ 76

dA/dx=8- 120/x^2=0

8=120/x^2

8x^2=120

x^2=15

x=sqrt(15) base

y=60/sqrt(15)=4sqrt(15) height

Area of the entire poster=(sqrt(15)+2)(4sqrt(15)+8)

=60+8sqrt(15)+8sqrt(15)+16=76+16sqrt(15…

Printed area = √15 x 4√15
Total area = (2+√15) x (8+4√15)

Let printed area be l x w, then A=(l+2)(w+8) and lw=60, so w=60/l

A=(l+2)(60/l+8)
A=60+8l+120/l+16
A=76+8l+120*l^-1
dA/dl=8-120*l^-2
For max/min dA/dl=0, so
0=8-120*l^-2
l=√15
w=60/√15=4√15