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?

1 Answer
Jun 10, 2015

Answer:

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:
enter image source here

#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)#.