OK, I need to find the minimum area that a triangle can occupy. The hypotenuse of the triangle passes through point (0, b) on the y axis, point [(5/2),(5/3)], and then point (a, 0) on the x axis. So I know that A=.5ab, and that I should set A’= 0 and solve. The problem is, I don’t know how to get a in terms of b so as to get one variable in the equation. I also have no idea what to do once I get that equation as it will have a variable and not the final area. Would I have to just plug in (5/2) if I get the equation in terms of a, being that a is an x variable (or 5/3 if I get the equation in terms of b)?

There is a formula for calculating the area of a triangle from knowing the length of the three sides, you might try that. It is called **Heron’s** formula… Since this formula uses the perimeter of the triangle, you could eaxmine the area as the perimeter approaches the length of the hypotenuse.

That might work but we don’t know the size of any of the sides, so I’m not so sure that it would help. I just thought, wouldn’ the smallest possible size of the hypotenuse be if the midpoint were [(5/2), (5/3)]? Could I use that to figure out anything?

Gah, not the length of the hypotenuse, but as the perimeter approaches twice the length of the hypoenuse. However, as I am working on it, it doesn’t seem to be going very well…

What I’m not sure I understand… does the hypotenuse stay the same length the entire time?

OK, if the midpoint forumula is useful, I got the midpoint= [(A+0/2), (B+0/2)]… then I set the (A/2)= 5/2 and (B/2)= 5/3

I ended up with A= 5 and B= 10/3… then multiplied them together to get the area. Please correct me if I’m wrong, which I probably am because I didn’t use derivatives. I’m also supposed to find the max area of the triangle, but I was taking it one step at a time…

A and B are variables…

You have the right answer with a=5, but I’m not sure your reasoning is correct (i.e. the minimum occurs when (5/2,5/3) is the midpoint of the hypotenuse). I’d have to think about it some more.

However, we do know that (a,0), (5/2,5/3) and (0,b) lie on a straight line. Thus the gradient of the line joining (a,0) to (0,b) is the same as that of the line joining (5/2,5/3) to (0,b). This gives us an equation connecting a & b. We can sustitute this into your area formula and then solve A’ = 0. This gives us your result of a=5.

I don’t get how you got the equation connecting a and b, could you please elaborate a little further?

In my head it seems that the area should approach zero… and that the minimum area of a triangle is, in fact, zero, when

(a+b)->c.

The minimum area of a triangle should be achieved as one of the variables, either a or b, approaches zero. The maximum area should be achieved when a=b at a->infinity.

For the minimum case, consider that b stays fixed and a approaches zero. In this case, ab/2 approaches zero trivially and thus the minimum area is zero.

For the maximum case, though, I am not confident that we can keep b fixed. Rather, I think they should be equal, as a*b is a maximum when a=b, and so the area of ab/2 should be a maximum when a=b. In this case we need to determine what the maximum length of a and b can be.

First, because a=b, the triangle is an isosceles. This means, most importantly, that A=B too. So we can essentially disregard angle a and split the triangle in half by drawing a line from ACB to c, that is, bisecting angle C, resulting in two right-angled triangles (hope you are following this) of equal area.

now we are left with a triangle with angle C’=C/2, c/2 as one leg, b’ as another leg (actually, the line which split our previous triangle in half) and a is now the hypotenuse.

This triangle is ALWAYS a right triangle since our b’ is always orthogonal to c. Also, its area is (1/2)(b’)(c/2)=(1/4)(b’c). Since c is constant, take the limit as b’->infinity, which results in an infinte area, which is what I would expect.

Thus, both halves of our original triangle reach infinite area, and so there is no maximum area defined by a triangle with only one side defined.

Someone else see this differently? I was hoping to work with the limits of the angles, because I am interested to know what the limit of angle C is for infinitely large triangles.

Anyone else see this differently?

In this case b=c, I should mention, which is the only way a can actually approach zero.

We know that (a,0), (5/2,5/3) and (0,b) lie on a stright line. The gradient of the line joining (a,0) to (0,b) is -b/a ( recall that gradient = change in y divided by change in x). The gradient of the line joining (a,0) to (5/2,5/3) is (5/3)/(5/2 - a), which equals 10/(15-6a). Since these three points lie on a line, these two gradients are equal.

Oh, shit, I get the problem, now. We are sort of rotating the hypotenuse around the point given in the op, not extending the legs to any old size. Sheesh. A little dense today. So the hypotenuse can grow and shrik, so long as it passess through this one point and makes a triangle with the two axis? Ok, got it.

Sorry 'bout that.

This looks to me like it should be in the General Questions forum, so I’m moving it over there.