The top and bottom margins of a poster are 8 cm and the side margins are each 6 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.
the top and bottom margins of a poster are 8 cm an
width = (include units)
height = (include units)

Dimensions of the poster

Width  29.10 cm

Height 38.80 cm

A(min) = 1129.08 cm²

Step-by-step explanation:

Printed area = 390 cm² = Ap

Lets call x and y dimensions of printed area

x   width

y   height

Then Ap = x*y    and

y = Ap/x    ⇒    y = 390/x

Then   total area of the poster is:

A(t)  = ( x  + 12 ) * ( y + 16 )         and   y = 390/x

A as a function of x

A(x) = ( x  + 12 ) * ( 390/x + 16 )   ⇒  A(x) = 390 + 16x + 4680/x + 192

A(x) = 582 + 16x + 4680/x     (1)

Taking derivatives

A´(x) =  16 - (4680/x²)         ⇒ A´(x)  =  0

[ 16x² -4680] /x² = 0     16x²- 4680 = 0    ⇒     x² = 4680/16

x = 17.10 cm        and    y = 390/x      y = 390/17.10      y = 22.80 cm

A(t) = ( 17.10 + 12 ) * ( 22.80 + 16 )

A(t) = 29.10 * 38.80

A(t) = 1129.08 cm²

If we substitute  in equation (1) the value of x  ( 17.10) we see A(x) > 0

Then there is a minimun at the point x = 17.10

can you give me more detail on this! ?

step-by-step explanation:

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