The top and bottom margins of a poster 6 cm each, and the side margins are 4 cm each. If the area of the printed material on the poster is fixed at 384 square centimeters, find the dimensions of the poster of smallest area. (list the smallest dimension first)

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jtellezd jtellezd · Answer 1 · 2019-11-22T16:48:11+01:00

Answer:

p = 25,20 cm

h = 34,32 cm

A(min) = 864,86 cm²

Step-by-step explanation:

Let call p and h dimensions of a poster, ( length , and height respectively) and x and y dimensions of the printed area of the poster then

p = x + 8 and h = y + 12

Printed area = A(p) = 384 cm² and A(p) = x*y ⇒ y = A(p)/x

y = 384 / x

Poster area = A(t) = ( x + 8 ) * ( y + 12 ) ⇒ A(t) = ( x + 8 ) * [( 384/x ) + 12 ]

A(t) = 384 + 12x + 3072/x + 96 A(t) = 480 + 3072/x + 12x

A(t) = [480x + 12x² + 3072 ] / x

A´(t) = [(480 + 24x )* x - (480x + 12x² + 3072]/x²

A´(t) = 0 [(480 + 24x )* x - 480x - 12x² - 3072] =0

480x + 24x² -480x -12x² - 3072 = 0

12x² = 3072 x² = 296

x = 17,20 cm and y = 384/17,20 y = 22,32 cm

Notice if you substitu the value of x = 17,20 in A(t) ; A(t) >0 so we have a minimun at that point

Then dimensions of the poster

p = 17,20 + 8 = 25,20 cm

h = 22.32 + 12 =34,32 cm

A(min) = 25,20 *34.32

A(min) = 864,86 cm²