Respuesta :
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
[tex]=xy[/tex]
we are given that area as 1536
so, we can set it to 1536
[tex]xy=1536[/tex]
now, we can solve for y
[tex]y=\frac{1536}{x}[/tex]
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
[tex]A=(8+x+8)\times (12+y+12)[/tex]
[tex]A=(x+16)\times (y+24)[/tex]
now, we can plug back y
[tex]A=(x+16)\times (\frac{1536}{x}+24)[/tex]
now, we have to minimize A
so, we will find derivative
[tex]A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)[/tex]
we can use product rule
[tex]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)[/tex]
[tex]=\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)[/tex]
now, we can simplify it
[tex]A'=-\frac{24576}{x^2}+24[/tex]
now, we can set it to 0
and then we can solve for x
[tex]A'=-\frac{24576}{x^2}+24=0[/tex]
[tex]-\frac{24576}{x^2}x^2+24x^2=0\cdot \:x^2[/tex]
[tex]-24576+24x^2=0[/tex]
[tex]x=32,\:x=-32[/tex]
Since, x is dimension
and dimension can never be negative
so, we will only consider positive value
[tex]x=32[/tex]
now, we can solve for y
[tex]y=\frac{1536}{32}[/tex]
[tex]y=48[/tex]
so, dimensions of printed poster are
length is 32 cm
width is 48 cm
We want to find the dimensions of the poster such that the area of the whole poster is minimized.
These are:
Length = 1560cm
Width = 17cm
We can assume that the poster is a rectangle, and we know that a rectangle of length L and width W has an area:
A = L*W
Here we do know that the top and bottom margins are 12cm each.
The side margins are 8cm each.
Then the measures of the printed areas are:
W' = W - 2*8cm = W - 16cm
L' = L - 2*12cm = L - 24cm
And we know that the area of the printed part is 1536 cm^2, then we can write:
A' = (W - 16cm)*( L - 24cm) = 1536 cm^2
W*L - 16cm*L - 24cm*W + 384cm^2 = 1536 cm^2
W*L = 16cm*L + 24cm*W + 1152cm^2
And W*L is equal to the area, so what we need to minimize is:
A = 16cm*L + 24cm*W + 1152cm^2
Here we can see that the dependence of the area is larger on W than on L, so what we need to minimize is W.
We will take the smallest value of W such that the given margins are allowed.
That value would be such that:
W - 16cm > 0
W > 16cm
this could (and to be strict, should) be something like 16.0001cm, but let's use 17cm just to use whole numbers, we will get:
W = 17cm
Then to get the length we solve:
W*L = 16cm*L + 24cm*W + 1152cm^2
17cm*L = 16cm*L + 24cm*17cm + 1152cm^2
1cm*L = 1560cm^2
L = 1560cm
These are the measures that minimize the area.
If you want to learn more, you can read:
https://brainly.com/question/10551873
