# Tough Problem

1. Jul 30, 2004

### Moose352

The line y=mx+b intersects the parabola y=x^2 at two points A and B. Find the point P on arc AOB that maximizes the area of triangle APB.

I want to see how you guys solve this problem. I wrote an equation for shortest distance between a point on the parabola and the line, then tried to maximize it. The problem is, I end up trying to find the zeros of a third degree polynomial. I may be screwing up the simple stuff, but I don't know.

2. Jul 30, 2004

### pnaj

Probably the thing to do is to derive an expression for the area of the triangle in terms of x, and then maximise that.

3. Jul 30, 2004

### HallsofIvy

Staff Emeritus
First clarify: does "arc AOB" mean the circular arc (most common meaning of arc) or the portion of the parabola from A to B?

4. Jul 30, 2004

### pnaj

That's a bit picky, isn't it?

5. Jul 30, 2004

### AKG

$$x_A = \frac{m + \sqrt{m^2 + 4b}}{2}$$

$$x_B = \frac{m - \sqrt{m^2 + 4b}}{2}$$

$$A(x_A, x_A ^2)\ \ B(x_B, x_B ^2)\ \ P(x_P, x_P ^2)$$

$$Area = \frac{1}{2} |\vec{AB} \times \vec{AP}|$$*

Express all this in terms of m, b, and $x_P$ and then differentiate area with respect to $x_P$.

Doing this, and setting the derivative to zero, I get:

$$x_P = \frac{x_A + x_B}{2} = \frac{m}{2}$$

Therefore:

$$P(\frac{m}{2}, \frac{m^2}{4})$$

*EDITED: "det(...)" to "|...|"

Last edited: Jul 30, 2004
6. Jul 30, 2004

### Moose352

Thanks AKG. But, how did you end up with the formula for the area? I thought the magnitude of the cross product is equal to the area of the triangle * 2. I don't understand how you have the determinant.

By the way, I figured out that I was making a stupid mistake differentiating, and I managed to find the answer that way too. But your way is a lot nicer.

And halls, AOB i guess means the portion of the parabola.

Moose

Last edited: Jul 30, 2004
7. Jul 30, 2004

### AKG

:) I imagine the problem would get very ugly if you didn't have that formula.

I can't remember the derivation, but the determinant of the cross product of two vectors gives the area of the parallelogram formed by those two vectors (of course, those vectors would be 2 of the 4 sides, but you should get the idea). The area of the triangle would just be half the area of that parallelogram. I'm at work right now, so I can't really try to figure out a derivation, but if you can't find one online I'll try to provide one (although I'm sure someone here will do that before I get a chance).

EDIT: Ignore most of the stuff above. I was a bit rusty with this stuff, so I looked it up on mathworld where their notation confused me, so what I wrote didn't make perfect sense. It's not the determinant of the cross product, but the magnitude of it which gives you the area of the parallelogram. If you have vectors $\vec{u} = (u_x, u_y)$ and $\vec{v} = (v_x, v_y)$ then the area of the parallelogram is:

$$A = |\vec{u} \times \vec{v}| = \det \left[\begin{array}{cc}u_x&u_y\\v_x&v_y\end{array}\right]$$

Now, how do we prove this?

Well, $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin \theta$, where $\theta$ is the angle between the vectors. Now, it's not hard to see that if the absolute value of the cross product is what's given on the right, that we have the area of the parallelogram. If $|\vec{u}|$ represents the base of the parallelogram, we can see with some trigonometry that $|\vec{v}|\sin \theta$ is the height/altitude of the parallelogram, and clearly the product of these two values is the area. Now, how do we know that the equation at the start of this paragraph is true? Well, I suppose it would be messy, but you'd just have to show that:

$$u_xv_y - v_xu_y = |\vec{u}||\vec{v}|\sin \theta$$

You can use the fact that:

$$\sin \theta = \sqrt{1 - \cos ^2 \theta} = \sqrt{1 - \left (\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \right )}$$

Last edited: Jul 30, 2004
8. Jul 30, 2004

### NateTG

The trick is that the cross product can also be written as a determinant, even if I don't understant the notation that AKG used (I thought the cross product was a vector...)

9. Jul 30, 2004

### Moose352

I understand it now! Thanks AKG.

10. Jul 30, 2004

### AKG

As I thought, the proof is "ugly". Well, not really, it's just rather involved, you can't be expected to just see it. Anyway, check out http://www.mcraeclan.com/MathHelp/GeometryVectorCrossProduct.htm [Broken] and look at the headings, "Magnitude of a x b" and the following heading "Area of a parallelogram".

Last edited by a moderator: May 1, 2017
11. Jul 30, 2004

### Gza

How's this for ugly: I did the problem with Heron's formula for the area of a triangle. I got the answer, but it took a good page of algebra. (Don't worry, I won't even think about posting it after seeing AKG's elegant solution.)

12. Jul 31, 2004

### uart

pnaj, Asking him to clarify the question is not being "picky". On the contrary, it's essentual. On reading the question I immediately wondered the same thing, was the "arc" an arc (three points define a unique cirular arc) or was the "arc" really the parabola. How can you do a problem that is not properly defined? I can't imagine how you could see that as being "picky".

13. Jul 31, 2004

### pnaj

It was just a comment, that's all.

I suppose that, technically, an arc refers to part of a circle, but I see the word 'arc' used in context with curves other than a circle all the time.

And I just didn't see the possible circle in this problem.

14. Jul 31, 2004

### HallsofIvy

Staff Emeritus
No, it wasn't a comment it was a question!

(NOW, I'm being picky!)

The only difference is that you assumed one possible meaning and I asked. I don't consider that "picky", just cautious.

15. Jul 31, 2004

### pnaj

HallsofIvy,

Apologies for my question.

pnaj