Mathematics
11.04.2021 08:31
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The top and bottom margins of a poster are each $3$ cm and the side margins are

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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meababy2009ow9ewa
meababy2009ow9ewa
5,0(40 marks)

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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redasher
redasher
4,6(50 marks)

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.

kingjonesjr
kingjonesjr
4,5(9 marks)
The answer is 0 because if you take x1 which is 4 and y1 which is 5 and subtract by x2 which is -3 and y2 which is 5 and reduce you’ll get 0

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