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.

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:

The area is given as 96 cm^2.

So, we have:

The print area is represented as:

Make x the subject in

Substitute 96/y for x in

Express as:

Expand

Differentiate

Set to 0

Add 4 to both sides

Multiply both sides by y^2

Divide both sides by 4

Take square roots

Rewrite as:

Recall that:

So, we have:

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

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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

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)  

   [ Putting  ]

Differentiating with respect to x

Again differentiating with respect to x

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

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

The width of the poster is = 8 cm

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

e.

step-by-step explanation:

because there are always 10 numbers in a phone number.

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