Calculas max/min problem: (-2. 4), (x, x^2), (4, 16) are three pts on curve y = x^2....

[pazzy78]

Hello all, I have been messing around with max/min problems and I am stuck on this one :

So I was thinking getting an expression for the area and trying to da/dx that would be too complex as there would be a lot of unknowns ...
So i tried to get the max of the lines (4,16) - (x,x^2) and (-2,4) - (x,x^2) .

Here is my attempts ..

can't be a coincidence that max/min vals for these are -2 and 4 ??


Anyway, was also thinking of getting an expression for Area by using 1/2 abSinC ... that would be great if I could only get the angle C ... I thought I had it when I extend the triangle as I know the slope of (-2,4) - (4,16) .. therefore the angle between that and the x axis is easy to get ... but the angle at the (4,16) vertex not so easy ...

Also using the 1/2 b*h ... OK I know the "base" .. but getting h is tough, since there is 2 unknowns in both points, it is tempting to use the diagram and say the point is on the y axis and the "base" line ... but I want a solid mathematical way of finding this.

I need direction here, any help appreciated.

 

[Dr.Peterson]

Hello all, I have been messing around with max/min problems and I am stuck on this one :

So I was thinking getting an expression for the area and trying to da/dx that would be too complex as there would be a lot of unknowns ...
So i tried to get the max of the lines (4,16) - (x,x^2) and (-2,4) - (x,x^2) .

Here is my attempts ..

can't be a coincidence that max/min vals for these are -2 and 4 ??

Do you mean you found the maximum length of each line? Why would that help find the maximum area??

What you found, of course, is that each line has a minimum length of 0, when x is at -2 or 4, so that the moving point is at one of the given points.

Anyway, was also thinking of getting an expression for Area by using 1/2 abSinC ... that would be great if I could only get the angle C ... I thought I had it when I extend the triangle as I know the slope of (-2,4) - (4,16) .. therefore the angle between that and the x axis is easy to get ... but the angle at the (4,16) vertex not so easy ...

Also using the 1/2 b*h ... OK I know the "base" .. but getting h is tough, since there is 2 unknowns in both points, it is tempting to use the diagram and say the point is on the y axis and the "base" line ... but I want a solid mathematical way of finding this.

You can just maximize h, since b is constant; there are several reasonable ways to calculate it. Give one a try. (Or at least tell us what you have learned that might be used. I assume you don't know vector methods?)

Note that it's easier if you maximize h^2, which avoids the square root.

No, the origin won't give the maximum area, if that's what your last comment means.

A valid "cheating" way is to find where the curve is parallel to the given chord; they probably want you to do it more directly than that.

 

[blamocur]

So I was thinking getting an expression for the area and trying to da/dx that would be too complex as there would be a lot of unknowns ...
So i tried to get the max of the lines (4,16) - (x,x^2) and (-2,4) - (x,x^2) .

Why do you think it would be too complex? Can you post your expression for the area in question?

I need direction here, any help appreciated.

Either a) get the expression for the area of the triangle, or b) use the fact that the (x,x^2) point should be the farthest point from the line ([-2,4], [4,16])

 

[pazzy78]

Got it out using the Area of large trapezium minus the sum of the other 2 ...

This is the correct answer..

 

[pazzy78]

Why do you think it would be too complex? Can you post your expression for the area in question?



Either a) get the expression for the area of the triangle, or b) use the fact that the (x,x^2) point should be the farthest point from the line ([-2,4], [4,16])

At first I was trying to use 1/2abSinC formula and that was really getting messy bringing in more unknowns.
Same with the trying to work out the perpendicular height.

But that trapezium method worked out nice, I was kicking myself as it was pointed out to me elsewhere ... can't believe I missed it..

 

[Dr.Peterson]

At first I was trying to use 1/2abSinC formula and that was really getting messy bringing in more unknowns.
Same with the trying to work out the perpendicular height.

But that trapezium method worked out nice, I was kicking myself as it was pointed out to me elsewhere ... can't believe I missed it..

You have done the equivalent of deriving a standard formula for the area of a triangle (or polygon) given the coordinates of its vertices. One version of this is called the Shoelace Formula:

In your case, the points are (4, 16), (-2, 4), and (x, x^2), so the area is [math]\frac{1}{2}\left((4)(4)-(-2)(16)+(x)(16)-(4)(x^2)+(-2)(x^2)-(x)(4)\right)=\frac{1}{2}\left(-6x^2+12x+48\right)=-3x^2+6x+24[/math]
I will admit I hadn't thought of this approach, but it is a good one. I expected you to find the foot of the perpendicular from the moving point to the fixed chord, and maximize that distance. A vector method would be easier if you had the tools, and would probably have a lot in common with the shoelace formula.

 

[pazzy78]

Very nice, I hadn't heard of that approach before.
I have another q I might post later that is difficult, but I have some ideas...



btw do all posts need to be pre approved ?

 

[stapel]

btw do all posts need to be pre approved ?

Just the first few posts need to be approved.