Mathematics, 30.03.2020 17:50 saifulcrc468

The top and bottom margins of a poster are each $3$ cm and the side margins are each $2$ cm. If the area of printed material on the poster is fixed at $96$ cm$^2$, find the dimensions of the poster with the smallest area.

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Answer from: 25linm

The dimension of the smallest area of the poster is 8 cm by 12 cm

Let the dimensions of the poster be x and y.

So, the area is:

$\mathbf{Area =xy}$

The area is given as 96 cm^2.

So, we have:

$\mathbf{xy = 96}$

The print area is represented as:

$\mathbf{A = (x - 4)(y - 6)}$

Make x the subject in $\mathbf{xy = 96}$

$\mathbf{x = \frac{96}{y}}$

Substitute 96/y for x in $\mathbf{A = (x - 4)(y - 6)}$

$\mathbf{A = (\frac{96}{y} - 4)(y - 6)}$

Express as:

$\mathbf{A = (96y^{-1} - 4)(y - 6)}$

Expand

$\mathbf{A = 96 -576y^{-1}- 4y + 24}$

Differentiate

$\mathbf{A' = 576y^{-2}- 4}$

Set to 0

$\mathbf{576y^{-2}- 4 = 0}$

Add 4 to both sides

$\mathbf{576y^{-2}= 4}$

Multiply both sides by y^2

$\mathbf{576= 4y^2}$

Divide both sides by 4

$\mathbf{144= y^2}$

Take square roots

$\mathbf{12= y}$

Rewrite as:

$\mathbf{y = 12}$

Recall that:

$\mathbf{x = \frac{96}{y}}$

So, we have:

$\mathbf{x = \frac{96}{12}}$

$\mathbf{x = 8}$

Hence, the dimension of the smallest area of the poster is 8 cm by 12 cm

Read more about areas at:

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Answer from: cupcake20019peehui

Therefore the dimension of the poster is 12 cm by 8 cm.

Step-by-step explanation:

Let the length of the poster be x and the width be y.

Given that the area of the poster is 96 cm².

∴xy =96

$\Rightarrow y= \frac{96}{x}$

The sides margins each are 2 cm and the top and bottom margins of the poster are each 3 cm.

The length of printing space is =(x- 2.3) cm

= (x-6) cm

The width of the printing space is =(y-2.2) cm

=( y-4 )cm

The area of the printing space is A=(x-6)(y-4) cm²

∴A=(x-6)(y-4)

$\Rightarrow A=(x-6)(\frac{96}{x}-4)$ [ Putting $y= \frac{96}{x}$ ]

$\Rightarrow A=120-\frac{576}{x}-4x$

Differentiating with respect to x

$A'= \frac{576}{x^2}-4$

Again differentiating with respect to x

$A''=-\frac{1152}{x^3}$

To find the minimum area, we set A'=0

$\therefore \frac{576}{x^2}-4=0$

$\Rightarrow \frac{576}{x^2}=4$

$\Rightarrow x^2=\frac{576}{4}$

$\Rightarrow x^2 =144$

$\Rightarrow x=\pm 12$

Dimension can't be negative.

Therefore x=12

If x=12, the value of A''>0,then at x=12, the area of the poster will be minimum.

If x=12, the value of A''<0,then at x=12, the area of the poster will be minimum.

$\therefore A''|_{x=12}=-\frac{1152}{12^3}$

Therefore at x= 12 cm the area of the poster will be maximum.

The width of the poster is $y=\frac{96}{12}$ = 8 cm

Therefore the dimension of the poster is 12 cm by 8 cm.

Answer from: Quest

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

Answer from: Quest

the length of o'p' is $1.5\times op$ .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

$k=\frac{o'p'}{op}$

the scale factor is 1.5.

$1.5=\frac{o'p'}{op}$

multiply both sides by op.

$1.5\times op=o'p'$

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is $1.5\times op$ .

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