Respuesta :
Answer:
The smallest poster has dimension 25.6 cm by 32.05 cm.
Step-by-step explanation:
Let "x" and "y" be the length and the width of the poster.
The margin of a poster are 4 cm and the side margins are 5 cm.
The length of the print = x - 2(4) = x - 8
The width of the print = y - 2(5) = y - 10
The area of the print = (x- 8)(y -10)
The area of the print is given as 388 square inches.
(x-8)(y -10) = 388
From this let's find y.
y -10 = [tex]\frac{388}{(x - 8)}[/tex]
y = [tex]\frac{388}{x - 8} + 10[/tex] -------------------(1)
The area of the poster = xy
Now replace y by [tex]\frac{388}{x - 8} + 10[/tex], we get
The area of the poster = x ([tex]\frac{388}{x - 8} + 10[/tex])
= [tex]10x + \frac{388x}{x - 8}[/tex]
To minimizing the area of the poster, take the derivative.
A'(x) = [tex]10 + 388(\frac{-8}{(x-8)^{2} } )[/tex]
A'(x) = [tex]10 - \frac{3104}{(x-8)^2}[/tex]
Now set the derivative equal to zero and find the critical point.
A'(x) = 0
[tex]10 - \frac{3104}{(x-8)^2}[/tex] = 0
[tex]10 = \frac{3104}{(x-8)^2}[/tex]
[tex](x - 8)^2 = \frac{3104}{10}[/tex]
[tex](x - 8)^2 = 310.4[/tex]
Taking square root on both sides, we get
x - 8 = 17.6
x = 17.6 + 8
x = 25.6
So, x = 25.6 cm takes the minimum.
Now let's find y.
Plug in x = 25.6 cm in equation (1)
y = [tex]\frac{388}{25.6 - 8} + 10[/tex]
y = 22.05 + 10
y = 32.05
Therefore, the smallest poster has dimension 25.6 cm by 32.05 cm.
