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?

Jun 10, 2015

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").
$A = 384 + 2 \left(\textcolor{red}{a \cdot 4}\right) + 2 \left(\textcolor{b l u e}{b \cdot 6}\right) - 4 \left(\textcolor{\mathmr{and} a n \ge}{6 \cdot 4}\right) = 384 + 8 a + 12 b - 96 = 288 + 8 a + 12 b$
As the sum of the illustrated parts:

$A = a \cdot b$
So let's get a in function of b:
$288 + 8 a + 12 b = a b$
$288 + 12 b = a b - 8 a$
$288 + 12 b = a \left(b - 8\right)$
$a = \frac{288 + 12 b}{b - 8}$

Now, $A \left(b\right)$ will be the function in one single variable (b) that we will minimize:
$A \left(b\right) = a \cdot b = \left(\frac{288 + 12 b}{b - 8}\right) \cdot b = \frac{288 b + 12 {b}^{2}}{b - 8} .$

I have to find the first derivative of the function to minimize it:
$A ' \left(b\right) = 12 \frac{{b}^{2} - 16 b - 192}{b - 8} ^ 2$
The minimum points satisfy the condition $A ' \left(b\right) = 0$, so:
${b}^{2} - 16 b - 192 = 0$
$b = - 8 \mathmr{and} b = 24$
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:
$a = \frac{288 + 12 b}{b - 8} = \frac{288 + 12 \cdot 24}{24 - 8} = 36$
So, the solution to the problem is $\left(a , b\right) = \left(24 , 36\right)$.