Aposter of area 24,000 cm2 has blank margins of width 10 cm on the top and bottom and 6 cm on the sides. find the dimensions that maximize the printed area. (let w be the width of the poster, and let h be the height.)

The total poster area is:

A = (l + 20) (w + 12)

where l and w is the length and width of the printed area

24000 = (l + 20) (w + 12)

l = [24000/(w + 12)] – 20

The printed area is simply:

a = l w

substituting l:

a = ([24000/(w + 12)] – 20) w

a = (24000w)(w + 12)^-1 – 20 w

Taking the first derivative da/dw:

da/dw = (24000)(w + 12)^-1 - (24000w)(w + 12)^-2 – 20

Set da/dw = 0:

(24000)/(w + 12) - (24000w)/(w + 12)^2 – 20 = 0

Multiply everything by (w + 12)^2:

24000 (w + 12) – 24000w - 20(w + 12)^2 = 0

24000w + 288000 – 24000w – 20 (w^2 + 24w + 144) = 0

-20 w^2 – 480 w + 285120 = 0

w^2 + 24 w = 14256

Completing the square:

(w + 12)^2 = 14256 + 12^2

w + 12 = ±120

w = 108 cm

l = [24000/(w + 12)] – 20

l = [24000/(108 + 12)] – 20

l = 180 cm

Therefore the dimensions of the whole poster are:

w + 12 = 120 cm

l + 20 = 200 cm

The poster should be 200 cm x 120 cm

x+y=90

x=30-y=x

step-by-step explanation:

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