Optimization Problem Finding Minimum Area

Question

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. If the area of printed material on the poster is fixed at 1536cm2, find the dimensions of the poster with the smallest area.

**Width = 48, but height does NOT = 32**

Answer 1

rektangel: aea=1536=x*y, so y=1536/x

tu get inside rektangel after kut off 12 at top & bottom

& kut off 8 from both sides

inside area=(x-16)*(y-24)

area=(x-16)*[(1536/x) -24]

look at x*area...z=(x-16)*(1536-24x)

-24x^2 +1536x +384x -24576

or   -24x^2 +1920x -24576

or +24x^2 -1920x +24,576

z=x^2-80x +1024

tu find min or max, yu wqnt dz/dz=0

dz/dz=2x -80=0, so 2x=80 or x=40

y=1536/x=1536/40=38.4

inside rektangel...y-24=14.4

x=40-16=24

inside area=x*y=14.4*24=345.6

Answer 2

The poster will look like that below.

printed material area =  hw = 1536   ------ (1)

Poster area is given by A = (w+16)(h+24)

A = wh + 24w + 16h + 384

A = (1536 + 384) + 24w + 16h    using (1)

A = 1920 + 24w + 16*1536/w      using (1) again

A = 1920 + 24w + 24576/w

Now differentiate A wrt w and set that to zero in order to determine the minimum of the function A(w).

dA/dw = 0 + 24 - 24567/w^2 = 0

1 - 1024/w^2 = 0

w^2 = 1024

w = 32,   h = 48

W = w + 16 = 48,    H = h + 24 = 72

Answer: Width = 48 cm,   Height = 72 cm