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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

A line passes through the point (1,2). What's the slope of the line passing through this point that would create a triangle of the least area. The x and y axis being the legs of a right triangle.

I went through what has to be done in my head but I don't know how to do it on paper.

I need to make a function that gives the area of the traingles based off m, find when the derivative equals 0 and find which gives a smaller output.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Start with the basics.

y = mx + b

We know this has to pass through the point (x, y) = (1, 2), so lets plug those in:

2 = m + b, or, m = 2 - b

That didn't seem to get us very far, did it? But at least we have a relationship for m and b, this might come in handy later.

So lets go back. What we want is a general equation for the area of the triangle. Well, we know of the formula 1/2 * base * height. So lets try to find those variables.

The height of the triangle is going to be the y-intercept, b. Easy enough.

The base of the triangle is going to be the x intercept. Since y = mx + b, and y must be 0 (definition of the x-intercept) 0 = mx + b. We want to solve this for x, so that would be x = -b / m.

So the area of the triangle is 1/2 * b * -b/m, or -b^2 / 2m. But wait, isn't this going to be negative? Negative area? Nope, remember the line that we are drawing has a negative slope, so m is negative, making -b^2 / m positive.

So we want to find the least area of the function -b^2 / 2m. Huh, two variables, that's going to be pretty tricky without multi variable calculus. But wait, doesn't m = 2 - b? Told you that would come in handy. So -b^2 / 2 * (2 - b) is the area, or -b^2 / (4 - 2b).

Try to find the minimum for that function. This will tell you what b is, then you can find m because m = 2 - b.

Edit:

And for extra credit, what kind of triangle does this make?

*Last edited by Ricky (2005-12-15 16:39:47)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

The x intercept must be greater than one.

Call x intercept J.

Here is the equation for the area.

Just find the J above one that makes the expression the smallest:

I arrived at this by drawing a picture and setting two

similar triangles proportions equal.

They are:

Sorry explanation so poor.

*Last edited by John E. Franklin (2005-12-15 17:21:55)*

**igloo** **myrtilles** **fourmis**

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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

Thank you both for helping.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Slope = -2

X intercept = 2

Y intercept = 4

*Last edited by John E. Franklin (2005-12-15 17:30:58)*

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

John, I'm curious how you came up with that. Think you could show more work? I can't seem to find where that equation is coming from.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

I think what he may have done is made a triangle by drawing a line from the origin to the point and making a right triangle.

What I'm most curious is if this congruency is true at all points.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Click on picture to see it bigger.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Add 2 to both sides to solve for Yintercept.

Multiply both sides of equation by Xintercept so as to

solve for the area; you can divide by 2, if you want to

be exact.

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Ah, I see now. Guess no calculus is needed.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

Wow thats really ingenious. Thats cool.

I'll show my teacher your method.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Well I cheated after that and plotted the points with a tiny BASIC program.

And noted the minimum area was 4 at x=2

I'm rusty at calculus, but I want to learn it again years later.

So this is a product, so there is a rule for that.

What is the derivative of this?

dJ

*Last edited by John E. Franklin (2005-12-15 18:19:26)*

**igloo** **myrtilles** **fourmis**

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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

The derivative of...

would use the product rule.

f'g+g'f

+ J*(-1/(x+1)^2)

I don't know how to use th math script..

(x^2+2x+2)/(x+1)^2

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Okay, I get the product rule.

But what is that last equation you wrote?

Also, the math script isn't necessary unless you

like to type even more to get it to look cool.

As long as your understood.

But [m a t h]\frac{3}{4}[/m a t h] makes a 3/4 fraction.

You can google on LaTeX Math to learn it.

\int is integrate

_{x=0} makes x=0 a subscript, the underscore does that.

^{I'm tiny north east} makes writing a superscript.

I'm going to bed now, it's 1:45 am here. Bye.

*Last edited by John E. Franklin (2005-12-15 18:46:32)*

**igloo** **myrtilles** **fourmis**

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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

Last equation was the derivative after being simplified.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

oh. I'll examine tomorrow. bye.

**igloo** **myrtilles** **fourmis**

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