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 = ![\frac{388}{(x - 8)}](https://tex.z-dn.net/?f=%5Cfrac%7B388%7D%7B%28x%20-%208%29%7D)
y =
-------------------(1)
The area of the poster = xy
Now replace y by
, we get
The area of the poster = x (
)
= ![10x + \frac{388x}{x - 8}](https://tex.z-dn.net/?f=10x%20%2B%20%5Cfrac%7B388x%7D%7Bx%20-%208%7D)
To minimizing the area of the poster, take the derivative.
A'(x) = ![10 + 388(\frac{-8}{(x-8)^{2} } )](https://tex.z-dn.net/?f=10%20%2B%20388%28%5Cfrac%7B-8%7D%7B%28x-8%29%5E%7B2%7D%20%7D%20%29)
A'(x) = ![10 - \frac{3104}{(x-8)^2}](https://tex.z-dn.net/?f=10%20-%20%5Cfrac%7B3104%7D%7B%28x-8%29%5E2%7D)
Now set the derivative equal to zero and find the critical point.
A'(x) = 0
= 0
![10 = \frac{3104}{(x-8)^2}](https://tex.z-dn.net/?f=10%20%3D%20%5Cfrac%7B3104%7D%7B%28x-8%29%5E2%7D)
![(x - 8)^2 = \frac{3104}{10}](https://tex.z-dn.net/?f=%28x%20-%208%29%5E2%20%3D%20%5Cfrac%7B3104%7D%7B10%7D)
![(x - 8)^2 = 310.4](https://tex.z-dn.net/?f=%28x%20-%208%29%5E2%20%3D%20310.4)
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 =
y = 22.05 + 10
y = 32.05
Therefore, the smallest poster has dimension 25.6 cm by 32.05 cm.