Question

The top and bottom margins of a poster are 4 cm and the side margins are each 6 cm. If the area of printed material on the poster is fixed at 384 square centimeters, how do you find the dimensions of the poster with the smallest area?

Answer 1

Draft

Explanation:

Let a be the width of the poster and b the height.
Let A be the area of the poster to be minimized.
(In this explanation I will omit all the "cm").
`A=384+2(color(red)(a*4))+2(color(blue)(b*6))-4(color(orange)(6*4))=384+8a+12b-96=288+8a+12b`
As the sum of the illustrated parts:

`A=a*b`
So let's get a in function of b:
`288+8a+12b=ab`
`288+12b=ab-8a`
`288+12b=a(b-8)`
`a=(288+12b)/(b-8)`

Now, `A(b)` will be the function in one single variable (b) that we will minimize:
`A(b)=a*b=((288+12b)/(b-8))*b=(288b+12b^2)/(b-8).`

I have to find the first derivative of the function to minimize it:
`A'(b)=12(b^2-16b-192)/(b-8)^2`
The minimum points satisfy the condition `A'(b)=0`, so:
`b^2-16b-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=(288+12b)/(b-8)=(288+12*24)/(24-8)=36`
So, the solution to the problem is `(a,b)=(24,36)`.