# Thread: Maximizing Area.

1. ## Maximizing Area.

A wire 10 cm long is cut into two pieces, one of length x and the other of length 10 - x. Each piece is bent into the shape of a square.

a. Find a function that models the total area enclosed by the two squares.

b. Find the value of x that minimizes the total area of the two squares.

2. Hi there,

If the first piece is folded into a square each side will be $\frac{x}{4}$ cm. If the second piece is folded into a square each side will be $\frac{10-x}{4}$ cm.

So the total area of both the squares will be defined by the function:

$f(x)=\frac{x}{4}\times \frac{x}{4}+\frac{10-x}{4}\times \frac{10-x}{4}$

For b) do you know how to rearrange this function to turning point form, so you can find the minimum?

3. There is a picture illustrated in my book for the problem. The full length of the string is 10 cm. One box is visibly smaller than the other box and is labeled x, where as the longer box is labeled 10 - x.

how do I rearrange this function to turning point form, so that I can find the minimum?

4. To find whatever value of x minimizes surface area, you need:

$\frac{dy}{dx} = 0$

and to make sure that

$\frac{d^2y}{dx^2} \geq 0$

First step is to get the function, but I think Stroodle has it.

EDIT: Unless I'm overthinking this problem....

5. How do I plug Stroodle's answer into your formula?

6. $\frac{dy}{dx}$ is the first derivative.

$\frac{d^2y}{dx^2}$ is the second derivative.

Do you know to take the derivative?

7. How does that have anything to do with the area?

8. When you take the derivative, you're finding the slope of the tangent lines.

When the derivative = 0, you know you have a max or a min on the graph.

We're trying to find a minimum.

The second derivative tests concavity. If the second derivative is less than 0, the graph is concave down. It looks like one mc-donalds arc, so to speak. This means that at the TOP of that arc, is the maximum. We want a minimum. So the second derivative should be POSITIVE (so the graph is convace up - looks like a smile). The very bottom of that "smile" is the minimum. This is what I like to call the second derivative test, to be sure that what you found from the FIRST derivative is the max OR min.

9. Hey Stroodle, how do I rearrange this function to turning point form, so that I can find the minimum?

10. I think what he meant is that you multiply out your equation to get it in a polynomial form so that it's easy to take the derivative of. I'm not 100% sure (we'll see what he says) but I think that's what he meant.

11. You can do it without calculus though.

The function (from my post above) is:

$f(x)=\frac{1}{8}(x^2-10x+50)$

Converting this to turning point form (by completing the square):

$f(x)=\frac{1}{8}\left [ (x-5)^2+25\right ]$

$f(x)=\frac{1}{8}(x-5)^2+\frac{25}{8}$

From this can see that the minimum area $\frac{25}{8}\ cm^2$ is where $x=5\ cm$

12. Sorry, I thought we were in the calculus forum.

Anyway, remember what I said since it'll be important for you in calculus.

13. Sorry, I don't understand. How did you get x = 5?

14. If x=5,

(5-5) = 0

Any other value of x will yield a greater value than 25/8 because anything squared is positive.

15. Lord Darkin is correct.

But otherwise, from basic quadratic theory you should know that if the coefficient of $x^2$ is positive the function will have a minimum, and that when the function is arranged into the form $f(x)=a(x-b)^2+c$ the minimum occurs at $(b,c)$