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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.

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Answer:

A(min) = 711,60 cm²

L = 15,40 cm

D = 46,21 cm

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

User Brett Bender
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