The top and bottom margins of a poster are 6 cm and the side margins are each 6 cm. If the area of printed material on the poster is fixed at 380 square centimeters, find the dimensions of the poster with the smallest area.

the solution to the problem is  

( a , b ) = ( 24 , 35.75 )

Step-by-step explanation:

Let a be the width of the poster and b the height.

Let A be the area of the poster to be minimized.

A = 380+ 2 ( a ⋅ 4 ) + 2 ( b ⋅ 6 ) − 4 ( 6 ⋅ 4 )

= 3840+ 8 a+ 12 b − 96

= 284+ 8 a + 12 b

A = a ⋅ b  

So let's get a in function of b:

284 + 8 a + 12 b = a b

284 + 12 b = a b − 8 a

284 + 12 b = a ( b − 8 )

 a =284+ 12 b/ b − 8

Now,  

A ( b )

will be the function in one single variable (b) that we will minimize:

A ( b ) = a ⋅ b

= ( 284 + 12 b b − 8 ) ⋅ b = 284 b + 12 b ^3/ b − 8

I have to find the first derivative of the function to minimize it:

A ' ( b ) = 12 b^ 2 − 16 b − 192/ ( b − 8 )^ 2

The minimum points satisfy the condition  

A ' ( b ) = 0

so:

b 2 − 16 b − 192 = 0

b = − 8  or  b = 24

But for obvious reason (b is the height of a poster), b must be positive, so only b=24 is a correct solution.

Now we have to find a:

a = 284+ 12 b/ b − 8 = 284+ 12 ⋅ 24 /24 − 8 = 35.75

So, the solution to the problem is  

( a , b ) = ( 24 , 35.75 )

x/3-10=8

x/3=18

x=48