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forsale [732]
3 years ago
13

PLSSSSSSSSSSSSSS HELP <3

Mathematics
1 answer:
Komok [63]3 years ago
3 0

1. Given

2. AD is parallel to BC

3. ∠3 ≅ ∠2

4. transitive property of congruence

5. Definition of bisect

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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
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Answer:

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.

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