1. ## Modeling with a function

I have never been good with modeling equations and this one is bugging me. I've been trying to practice in my book but just can't seem to figure this one out. Would be much appreciated if someone could help.

A wire L = 14 cm long is cut into two pieces, one of length x and the other of length (L - x). Each piece is bent into the shape of a square.

Find a function that models the total area A enclosed by the two squares in terms of x

2. Are you doing it this way as well?

Since I get a positive quadratic, when I differentiate that I can only get the value of x for the smallest possible area.

Showcase_22=epic fail.

3. Wow, I forgot to square the 4 on the $(x/4)^2$. Thanks for the help. I don't understand how to minimize it either. I know to maximize it you can do $-b/2a$ but to minimize it they say I should graph it but I dunno how to graph that. Is there a shortcut to minimizing it?

4. Originally Posted by n8thatsme
Wow, I forgot to square the 4 on the $(x/4)^2$. Thanks for the help. I don't understand how to minimize it either. I know to maximize it you can do $-b/2a$ but to minimize it they say I should graph it but I dunno how to graph that. Is there a shortcut to minimizing it?
After substituting L = 14 and simplifying you get $A = \frac{x^2 - 14x + 98}{8}$. The domain is 0 < x < 14. This is a simple quadratic and you should have no trouble drawing its graph.

The minimum value of A occurs at the minimum turning point. The maximum value occurs at an endpoint.

5. That's what I was doing wrong!

I forgot that there was a limit for the maximum value of x.

I'll remember that for other mathy times.

6. So am I putting the numerator into standard form? This stuff just doesn't make any sense to me.

7. Originally Posted by n8thatsme
So am I putting the numerator into standard form? This stuff just doesn't make any sense to me.
You've been given the rule. It's a parabola. Draw the graph of this parabola in the way you've been taught.

If you're working on worded problems like this one you must have already studied parabolas and how to draw their graphs. You are stongly advised to go back and extensively revise this pre-requisite material.

8. Since the x^2 term in the quadratic equation is positive, you should get a graph that has a minimum point. This is the value of x for which the area will be smallest (and what I was getting confused on earlier).

Remembering that x is between 0 and 14cm, the maximum value of x can be found. I'll leave that up to you.

After you've found your value, it does help substituting numbers either side of your value to see if you really have found the solution. It helps me anyway!