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

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MarkFL

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Feb 24, 2012
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Here is the question:

OPTIMIZATION PROBLEM HELP!!! PLEASE?

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:

OPTIMIZATION PROBLEM HELP!!! PLEASE? - Yahoo! Answers

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

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Feb 24, 2012
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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
Jan 29, 2012
661
Congratulations Mark. One of the most elegant posts I've ever seen.