Word Problem

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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

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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:
[tex]A_{sq.}=(\frac{x}{4})*(\frac{x}{4})=\frac{x^{2}}{16}[/tex]

Agreed?

Now, you've got an equilateral triangle, whose side is [tex]s=\frac{10-x}{3}[/tex]
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 [tex]s,\frac{s}{2},h[/tex] where "h" is the height.
Our triangle is right-angled, so we have, by Pythogoras:
[tex]s^{2}=(\frac{s}{2})^{2}+h^{2}[/tex]
that is we get:
[tex]h=\frac{\sqrt{3}}{2}s[/tex]

Now, can you take it from here?
 
Last edited:
14
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how did you get

[tex]h=\frac{\sqrt{3}}{2}s[/tex]


from

[tex]s^{2}=(\frac{s}{2})^{2}+h^{2}[/tex]


??
 

arildno

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[tex]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[/tex]
 
14
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Yeah I should have seen that.

So now that I know that the height

[tex]h=\frac{\sqrt{3}}{2}s[/tex]


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

[tex]A=\frac{1}{2}(\frac{10-x}{6})(\frac{\sqrt{3}}{2})(\frac{10-x}{3})[/tex]

[tex]A=\frac{100\sqrt{3}-20x\sqrt{3}-x^2\sqrt{3}}{72}[/tex]

Now I add the to area formulas togther to get

[tex]A=\frac{100\sqrt{3}-20x\sqrt{3}-x^2\sqrt{3}}{72} + \frac{x^2}{4}[/tex]

Is this correct?
 

arildno

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

You should have for the total area A:
[tex]A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{10-x}{3})^{2}[/tex]
 
14
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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

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Have you learnt about differentiation (derivatives) yet?
 
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no not yet
 

arildno

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

Are you sure about that?
Haven't you covered minimization problems in class?
 
14
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in this

[tex]A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{10-x}{3})^{2}[/tex]

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

Just wondering.
 
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The half has already been taken into account, that's why you have [tex]\frac{\sqrt{3}}{4}[/tex] and not [tex]\frac{\sqrt{3}}{2}[/tex], ie [tex]\frac{\sqrt{3}}{2}\frac{1}{2} = \frac{\sqrt{3}}{4}[/tex].
 
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arildno

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You have, for the area of the triangle:
[tex]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}[/tex]
 
14
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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
 
14
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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

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

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