The top and bottom margins of a poster are 4 cm and the side margins are each 5 cm. if the area of printed material on the poster is fixed at 388 square centimeters, find the dimensions of the poster with the smallest area.

The smallest poster has dimension 25.6 cm by 32.05 cm.

Step-by-step explanation:

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

The margin of a poster are 4 cm and the side margins are 5 cm.

The length of the print = x - 2(4) = x - 8

The width of the print = y - 2(5) = y - 10

The area of the print = (x- 8)(y -10)

The area of the print is given as 388 square inches.

(x-8)(y -10) = 388

From this let's find y.

y -10 =

y = -------------------(1)

The area of the poster = xy

Now replace y by , we get

The area of the poster = x ()

=

To minimizing the area of the poster, take the derivative.

A'(x) =  

A'(x) =

Now set the derivative equal to zero and find the critical point.

A'(x) = 0

= 0

Taking square root on both sides, we get

x - 8 = 17.6

x = 17.6 + 8

x = 25.6

So, x = 25.6 cm takes the minimum.

Now let's find y.

Plug in x = 25.6 cm in equation (1)

y =  

y = 22.05 + 10

y = 32.05

Therefore, the smallest poster has dimension 25.6 cm by 32.05 cm.

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.