Answer:
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
