# tangoforever's question at Yahoo! Answers regarding minimizing area of poster

#### MarkFL

Staff member
Here is the question:

The top and bottom margins of a poster are 4 cm and the side margins are each 2 cm. If the area of printed material on the poster is fixed at 384 square centimeters, find the dimensions of the poster with the smallest area.
Here is a link to the question:

I have posted a link there to this topic so the OP can find my response.

#### MarkFL

Staff member
Re: tangoforever's question at Yahoo! Answers regarding minimzing area of poster

Hello tangoforever,

Let's let $h$ be the height of the poster, and $w$ be the width. All linear measures will be in cm.

We know the height of the printed area is $h-2\cdot4=h-8$ and the width of the printed area is $w-2\cdot2=w-4$. We are given the area of the printed area to be $384\text{ cm}^2$. And since the area of a rectangular region is width times height, we may state:

(1) $$\displaystyle 384=(w-4)(h-8)$$

Now, the area of the poster, which we will denote by $A$, is simply width times height, or:

(2) $$\displaystyle A=wh$$

This is the quantity we wish to minimize. Using (1), we may solve for either variable, and then substitute into (2) to get a function in one variable. Let's replace $w$, and so solving (1) for $w$, we find:

(3) $$\displaystyle w=\frac{384}{h-8}+4$$

At this point we may want to recognize that we require $8<h$.

And so we find:

$$\displaystyle A(h)=\left(\frac{384}{h-8}+4 \right)h$$

Next we want to equate the first derivative to zero to find the critical values in the domain of the function. Using the product rule for differentiation, we find:

$$\displaystyle A'(h)=\left(\frac{384}{h-8}+4 \right)(1)+\left(-\frac{384}{(h-8)^2} \right)h=\frac{384(h-8)+4(h-8)^2-384h}{(h-8)^2}=\frac{4\left(h^2-16h-704 \right)}{(h-8)^2}=0$$

Application of the quadratic formula, and discarding the root outside of the domain, yields the critical value:

$$\displaystyle h=8+16\sqrt{3}=8(1+2\sqrt{3})$$

Use of the first derivative test shows that the derivative is negative to the left of this critical value and positive to the right, hence this critical value is at a local minimum, and in fact is the global minimum on the restricted domain.

Here is a plot of the area function for $$\displaystyle 8\le h\le64$$:

Now, to find the width, we may use (3) which yields:

$$\displaystyle w=\frac{384}{8(1+2\sqrt{3})-8}+4=4(1+2\sqrt{3})=\frac{h}{2}$$

To tangoforever and any other guests viewing this topic, I invite and encourage you to post other optimization questions in our Calculus forum.

Best Regards,

Mark.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Congratulations Mark. One of the most elegant posts I've ever seen.