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weqwewe [10]
3 years ago
11

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 po

ster is fixed at 1536 cm2. Find the dimensions of the printed area that minimize the area of the whole poster.
Mathematics
1 answer:
grin007 [14]3 years ago
5 0

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


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