The dimensions of the poster be so that the least amount of poster is used are 10 and 20 inches.
Print area of the rectangular poster =128 in²
Let "x" and "y" be the dimensions for the print area of the poster .
And A[p] = print area of poster
⇒A(p) = Print area of the poster = xy
⇒128 = xy
⇒ y = 128/x
Total area of the poster is:
A( t ) = ( y + 4 )( x + 2 )
A( t ) = yx +2y +4x + 8 And as y = 128/x
Poster 's area as function of x is -
A(x) =( 128 /x)x + 2 (128/x) + 4x + 8
⇒ A(x) = 128 x² + 256/x + 4x + 8 ⇒ A(x) = 128 + 256 /x + 4*x
On taking the derivatives on both sides of the equation :
A´(x) = - 256/x² + 4
A´(x) = 0
⇒ - 256 /x² = -4
⇒ 4x² = 256
⇒ x² = 256/4 ⇒ x² = 64
x = 8inches
And y = 128/x
⇒y = 128/8 ⇒ y = 16 inches
After finding out the values of x and y the dimensions of the poster are:
w = x + 2 ⇒ w = 8+ 2 w = 10 in
Y = y + 4 ⇒ Y = 16+ 4 Y = 20 in
Hence, the dimensions of the poster be so that the least amount of poster is used are 10 and 20 inches.
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