Question

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. The area of printed material on the poster is fixed at 1536 cm2. Find the dimensions of the printed area that minimize the area of the whole poster.

Answer 1

Answer:

Dimensions of printed poster are

length is 32 cm

width is 48 cm

Step-by-step explanation:

Let's assume

length of printed poster is x cm

width of printed poster is y cm

now, we can find area of printed poster

so, area of printed poster is

$=xy$

we are given that area as 1536

so, we can set it to 1536

$xy=1536$

now, we can solve for y

$y=\frac{1536}{x}$

now, we are given

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm

so, total area of poster is

$A=(8+x+8)\times (12+y+12)$

$A=(x+16)\times (y+24)$

now, we can plug back y

$A=(x+16)\times (\frac{1536}{x}+24)$

now, we have to minimize A

so, we will find derivative

$A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)$

we can use product rule

$A'=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

$=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

now, we can simplify it

$A'=-\frac{24576}{x^2}+24$

now, we can set it to 0

and then we can solve for x

$A'=-\frac{24576}{x^2}+24=0$

$-\frac{24576}{x^2}x^2+24x^2=0\cdot \:x^2$

$-24576+24x^2=0$

$x=32,\:x=-32$

Since, x is dimension

and dimension can never be negative

so, we will only consider positive value

$x=32$

now, we can solve for y

$y=\frac{1536}{32}$

$y=48$

so, dimensions of printed poster are

length is 32 cm

width is 48 cm