The top and bottom margins of a poster are 6 cm and the side margins are each 2 cm. if the area of printed material on the poster is fixed at 390 square centimeters, find the dimensions of the poster with the smallest area.

A(min)  =   711,60  cm²

L  = 15,40  cm

D  = 46,21 cm

Step-by-step explanation:

Let L   and  D  the lenght   and height of the poster

and   x     and    y     dimensions of printed area so

L   =   y  + 12        and        D   =   x   + 4

Printed area   =  390 cm²       and        Pa  =   x*y     then     y  =  390/x

Poster area is    Ap   =   L * D      =   (y  +  12 )  *  ( x  +  4 )

A(x)   =  ( 390/x   +  12  )  *  (  x  +  4  )

A(x)   =  390 + 1560/x   + 12x  + 48

A(x)   = 438   +  1560/x  + 12x

Taking derivatives on both sides of the equation

A´(x)   =   12  - 1560/x²

A´(x)  =  0            12  - 1560/x²  =  0      12x²  =  1560

x²  =  √ 130

x    =    11.40  cm       then    y  390/ 11,40         y  =  34,21  cm

And    L  =  11,40  +  4     L  =  15,40 cm      

And    D  =  34,21  + 12    D =  46,21  cm

A(min)  =   15,40* 46,21

A(min)  =   711,60  cm²

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

my guess is

step-by-step explanation: