A rectangular poster is to contain 882 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the poster be so that the least amount of poster is used

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

1. 34

2. 268

3. x=14.4

step-by-step explanation: