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

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answer:
congruent segments
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Answer from: Quest

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