# Word Problem

1. Jun 12, 2005

### Mathnewbie

Hello I have this word problem that I am having problems solving. Hopefully someone can help

Here is it.

A piece of wire 10m long is cut into two pieces. On piece, of length x, is bent into the shape of a square. The other is bent into the shape of a equilateral triangle.

(a) expess the total area of a enclosed as a function of x
b. For what value is of x is this total area a minimum.

This is what I think I can figure out.

One side of the square is s = x/4
One side of the triangle is s = (10 -x)/3

Area of Square is A= lw
Area of Triangle is A = 1/2bh

How do I go from here.

TIA

I may be totally wrong

2. Jun 12, 2005

### arildno

You're on the right track!
Let's take the area of the square first.
Since the length is equal to the width for the square, we have that the square's area is:
$$A_{sq.}=(\frac{x}{4})*(\frac{x}{4})=\frac{x^{2}}{16}$$

Agreed?

Now, you've got an equilateral triangle, whose side is $$s=\frac{10-x}{3}$$
What you need to figure out, is the height of that triangle.
To do that, look on the triangle formed by dropping a perpendicular from one vertex down on one side (dividing that side in half), whose side-lengths must be $$s,\frac{s}{2},h$$ where "h" is the height.
Our triangle is right-angled, so we have, by Pythogoras:
$$s^{2}=(\frac{s}{2})^{2}+h^{2}$$
that is we get:
$$h=\frac{\sqrt{3}}{2}s$$

Now, can you take it from here?

Last edited: Jun 12, 2005
3. Jun 12, 2005

### Mathnewbie

how did you get

$$h=\frac{\sqrt{3}}{2}s$$

from

$$s^{2}=(\frac{s}{2})^{2}+h^{2}$$

??

4. Jun 12, 2005

### arildno

$$h^{2}=s^{2}-\frac{s^{2}}{4}=\frac{4s^{2}-s^{2}}{4}=\frac{3}{4}s^{2}\to{h}=\frac{\sqrt{3}}{2}s$$

5. Jun 12, 2005

### Mathnewbie

Yeah I should have seen that.

So now that I know that the height

$$h=\frac{\sqrt{3}}{2}s$$

I can plug s into the height formula and the height formula into the area like this

$$A=\frac{1}{2}(\frac{10-x}{6})(\frac{\sqrt{3}}{2})(\frac{10-x}{3})$$

$$A=\frac{100\sqrt{3}-20x\sqrt{3}-x^2\sqrt{3}}{72}$$

Now I add the to area formulas togther to get

$$A=\frac{100\sqrt{3}-20x\sqrt{3}-x^2\sqrt{3}}{72} + \frac{x^2}{4}$$

Is this correct?

6. Jun 12, 2005

### arildno

Basically, yes,; but you've made a couple of arithmetical mistakes.

You should have for the total area A:
$$A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{10-x}{3})^{2}$$

7. Jun 12, 2005

### Mathnewbie

cool thank i will try to figure out where made the mistakes.

how do i figure out this part

For what value is of x is this total area a minimum

Thanks again

8. Jun 12, 2005

### arildno

Have you learnt about differentiation (derivatives) yet?

9. Jun 12, 2005

### Mathnewbie

no not yet

10. Jun 12, 2005

### arildno

And you were given this as an assignment all the same??

Haven't you covered minimization problems in class?

11. Jun 12, 2005

### Mathnewbie

in this

$$A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{10-x}{3})^{2}$$

what happen 1/2 from the area of triangle??

Just wondering.

12. Jun 12, 2005

### Nylex

The half has already been taken into account, that's why you have $$\frac{\sqrt{3}}{4}$$ and not $$\frac{\sqrt{3}}{2}$$, ie $$\frac{\sqrt{3}}{2}\frac{1}{2} = \frac{\sqrt{3}}{4}$$.

Last edited: Jun 12, 2005
13. Jun 12, 2005

### arildno

You have, for the area of the triangle:
$$A_{tr}=\frac{1}{2}*h*s=\frac{1}{2}*\frac{\sqrt{3}}{2}*s^{2}=\frac{\sqrt{3}}{4}s^{2},s=\frac{10-x}{3}$$

14. Jun 12, 2005

### Mathnewbie

all we have done with min or max is in the for of ax^2 + bx + c

if a < 0 then x= -(b)/2a is the max
if a > 0 then x= -(b)/2a is the min

this was a assignment and I got it wrong and think there will be something like it on the test. This is the only min max question we have had.

15. Jun 12, 2005

### Mathnewbie

I think I need to be a little more organized when I am solving a question so I know what I have done already. Thanks for the help.

16. Jun 12, 2005

### arildno

Yes, then what you need to do is to identify "a" and "b" in your expression for A(x)!