Solving Word Problem: 10m Wire Cut into Square & Triangle

[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

 

[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?

 

[Mathnewbie]

how did you get

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


from

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


??

 

[arildno]

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

 

[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?

 

[arildno]

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

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

 

[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

 

[arildno]

Have you learned about differentiation (derivatives) yet?

 

[Mathnewbie]

no not yet

 

[arildno]

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

Are you sure about that?
Haven't you covered minimization problems in class?

 

[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.

 

[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 $$\frac{\sqrt{3}}{2}\frac{1}{2} = \frac{\sqrt{3}}{4}$$.

 

[arildno]

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

 

[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.

thanks for all your help

 

[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.

 

[arildno]

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