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) $$384=(w-4)(h-8)$$
Now, the area of the poster, which we will denote by $A$, is simply width times height, or:
(2) $$A=wh$$
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) $$w=\frac{384}{h-8}+4$$
At this point we may want to recognize that we require $8<h$.
And so we find:
$$A(h)=\left(\frac{384}{h-8}+4 \right)h$$
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:
$$A'(h)=\left(\frac{384}{h-8}+4 \right)(1)+\left(-\frac{384}{(h-8)^2} \right)h=\frac{384(h-8)+4(h-8)^2-384h}{(h-8)^2}=\frac{4\left(h^2-16h-704 \right)}{(h-8)^2}=0$$
Application of the quadratic formula, and discarding the root outside of the domain, yields the critical value:
$$h=8+16\sqrt{3}=8(1+2\sqrt{3})$$
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 $$8\le h\le64$$:
Now, to find the width, we may use (3) which yields:
$$w=\frac{384}{8(1+2\sqrt{3})-8}+4=4(1+2\sqrt{3})=\frac{h}{2}$$
To tangoforever and any other guests viewing this topic, I invite and encourage you to post other optimization questions in our http://www.mathhelpboards.com/f10/ forum.
Best Regards,
Mark.