the top and bottom margins of a poster are 4 cm and the side margins are each 2 cm. If the area of printed materail on the poster is fixed at 380 cm^2, find the dimensions of the poster with teh smallest area

The dimension of the poster are

length: √190 + 4 cm

heigth: √760 + 8 cm

Step-by-step explanation:

Let X be the length of the printed area and Y be the heigth. Then X*Y = 380. Note that Y = 380/X.

The length of the poster is x+2*2 = x+4 and the heigth is Y + 4*2 = Y+8 = 380/X + 8. Thus, the area of the poster is

f (X) = (X+4) * (380/X + 8) = 380 + 8X + 1520/X + 32 = 8x + 412 + 1520/X.

We want to minimize f (X), so we derivate it.

f' (x) = 8 + 0 - 1520/X² = 8-1520/X².

The derivate is equal to 0 when

8 = 1520/X²

X² = 1440/8 = 190

X = √190.

Thus Y = 380/√190 = √760.

The dimensions of the poster with the smallest area are:

length: √190 + 4 cm

heigth: √760 + 8 cm