Triangles/Max-Min Question (1 Viewer)

[Amleops]

Consider all triangles formed by lines passing through the point (8/27, 9) and both the x and y axis.

a) Show that the length of the hypotenuse (h) is h = sqrt(x^2 + (9x/(x-(8/27)))^2)
b) Find the value of x that will give the shortest hypotenuse

For Part a, having deduced h = sqrt(x^2 + y^2) I tried to find y through the point-gradient formula (subbing in 8/27, 9) which yielded no results, or if it did, there must have been a mistake in my algebra. I had also started on using a method involving similar triangles but I thought I was making too many assumptions so I didn't pursue that any further.

For Part b, having differentiated h I got (2x+(9(x-(8/27))-9x)/(x-(8/27))^2) / 2sqrt(x^2 + 9x/(x-(8/27)) which was quite ugly, equating to zero I eventually got something which resembled an equation reducible to quadratics except it had an extra variable in there which was difficult to do anything with.

Any suggestions?

 

[Fus Ro Dah]

Ummm.... the shortest hypotenuse is simply h=0 as per the case when the equation of the line is 8y=243x.

Did you mean the positive x and positive y axis?

 

[Amleops]

Did you mean the positive x and positive y axis?

The question didn't say anything specifying otherwise. But I think that is implied, yes.

As for the hypotenuse I'm pretty sure it can't equal zero; Part B is worth 4 marks so it don't think it's that simple.

 

[Fus Ro Dah]

The question didn't say anything specifying otherwise. But I think that is implied, yes.

As for the hypotenuse I'm pretty sure it can't equal zero; Part B is worth 4 marks so it don't think it's that simple.

Yes it can be equal to zero. Please refer to my example of the case where the line is in the form y=mx and passing through the given point. I answered the question according to what was given, and there was no such specification that x>0 and y>0. If I got this question in an exam and the marker gave me zero marks for giving the answer "h=0" along with my example, then I will fight to the death for those 4 marks.

As for the question, note that m<0 and that m is a function of x and y.

 

[Amleops]

Yeah when I was using the point-gradient formula I had m = -y/x where m<0 and m is a function of x and y as you have said, after expanding I couldn't get the answer, so it might have been due to an algebraic error if that method was correct.

The only thing I forgot to add from this question was a given diagram with an example of two such triangles where x>0 and y>0, so I was implying from that. That being said, the question makes no reference to the possibility that the triangles could be in the negative axes, so I can understand the ambiguity. Sorry for the confusion.

 

[Fus Ro Dah]

Yeah when I was using the point-gradient formula I had m = -y/x where m<0 and m is a function of x and y as you have said, after expanding I couldn't get the answer, so it might have been due to an algebraic error if that method was correct.

The only thing I forgot to add from this question was a given diagram with an example of two such triangles where x>0 and y>0, so I was implying from that. That being said, the question makes no reference to the possibility that the triangles could be in the negative axes, so I can understand the ambiguity. Sorry for the confusion.

No worries

Now your function for m in terms of x and y is actually incorrect. What you have there implies that the line goes through the origin, but that is not necessarily the case.

 

[Amleops]

OK.....but if you sub in (0,0) to the gradient you get an undefined answer, so wouldn't that mean it doesn't pass through the origin? I don't know, what gradient did you get?

 

[Fus Ro Dah]

The Gradient formula is for some constants x_1 and y_1, which we will let be the coordinate given. The case when it passes through the origin contradicts what I said before about m<0.

 

[RealiseNothing]

Ummm.... the shortest hypotenuse is simply h=0 as per the case when the equation of the line is 8y=243x.

Did you mean the positive x and positive y axis?

But by definition that wouldn't be classified as a triangle unless the question stated otherwise, so h=0 is not included.

 

[Fus Ro Dah]

But by definition that wouldn't be classified as a triangle unless the question stated otherwise, so h=0 is not included.

http://en.wikipedia.org/wiki/Degeneracy_(mathematics)

 

[RealiseNothing]

http://en.wikipedia.org/wiki/Degeneracy_(mathematics)

Within context however.

 

[barbernator]

for a triangle to be formed with the positive x axis, and the positive y axis, the gradient has to be negative does it not? if fus is assuming the point at the origin is a degenerate triangle, that is not really inclusive of the line anyways

 

[Fus Ro Dah]

for a triangle to be formed with the positive x axis, and the positive y axis, the gradient has to be negative does it not?

As for the question, note that m<0 and that m is a function of x and y.

Yeah

 

[barbernator]

Yeah

sorry, didnt read the rest of your reasining

 

[Amleops]

I think I've got an answer for b, but it was long and tedious, and I had to use a bit of guesswork and a graphing calculator for the last bit (although I could have found it myself using calculus but it would've taken much longer).
































There's definitely a better way to do this though, any suggestions?