Area of the largest triangle

[MathsLearner]

I am trying to solve the problem

I am not very confident of this problem, i just made an attempt for the part a. The diagram will look like this

The equations of the green and blue lines are
[math] y = -x +1; \\ y = x +1; \\ Area = \int_{-1}^1(-x+1){dx} \\ [x - \frac {x^2} 2]_{-1}^1 \\ 2 - 1 = 1; \\ A=1 [/math]Is the method correct? Please advise

 

[HallsofIvy]

Is there a reason for integrating to find the area of a triangle? Your triangle has base length 2 and height 1 so area (1/2)(2)(1)= 1.
Second, having said that one side of the triangle, for x from -1 to 0, is y= x+ 1, and that the other, for x from 0 to 1, is y= -x+ 1, why do you integrate -x+ 1 from -1 to 1? Shouldn't you integrate x+ 1 from -1 to 0 and -x+ 1 from 0 to 1? Indeed, you have \(\displaystyle \left[x- \frac{x^2}{2}\right]_{-1}^1\) which is actually (1- 1/2)- (-1- 1/2)= 1/2+ 3/2= 2 but then you write 2- 1= 1! Which just happens to be the correct answer!

The correct integral for the triangle you have is would be \(\displaystyle \int_{-1}^0 x+ 1 dx+ \int_0^1 -x+ 1 dx= \left[\frac{x^2}{2}+ x\right]_{-1}^0+ \left[-\frac{x^2}{2}+ x\right]_0^1= (0- \frac{1}{2}+ 1)+ (-\frac{1}{2}+ 1)= \frac{1}{2}+ \frac{1}{2}= 1\).

Also, how do you know that this triangle is the largest triangle that will fit into that circle? What about the equilateral triangle with vertices equally spaced around the circle?

Finally, part (b) says "inside that parabola". What parabola?

 

[JeffM]

Well I am not sure. The problem does not mention lines. It seems to me that you simply assumed what I admit is intuitive, namely that the largest triangle will have a base of 2 and a height of 1. So clearly the area of that triangle by a formula learned years ago is

[MATH]\dfrac{1}{2} * bh \implies \dfrac{1}{2} * 2 * 1 = 1.[/MATH]
So you can find the correct answer using integrals, but you do not need such a high powered tool. However, how did you prove that the triangle with the largest area is the one you drew? That seems to me to be a problem in differential calculus subject to constraints.

Without context, I have no clue whether you used the "correct" method.

 

[lookagain]

Also, how do you know that this triangle is the largest triangle that will fit into that circle? What about the equilateral triangle with vertices equally spaced around the circle?

This is not about the area of the largest triangle that will
fit into that circle. The base of the triangle must lie on
the x-axis as a restriction for part (a).

 

[MathsLearner]

Is there a reason for integrating to find the area of a triangle? Your triangle has base length 2 and height 1 so area (1/2)(2)(1)= 1.

I just wanted to get an equation in terms of the variable x. In general to find max or min we use the differentiation, i was trying if i can get one to differentiate.

Second, having said that one side of the triangle, for x from -1 to 0, is y= x+ 1, and that the other, for x from 0 to 1, is y= -x+ 1, why do you integrate -x+ 1 from -1 to 1? Shouldn't you integrate x+ 1 from -1 to 0 and -x+ 1 from 0 to 1? Indeed, you have \(\displaystyle \left[x- \frac{x^2}{2}\right]_{-1}^1\) which is actually (1- 1/2)- (-1- 1/2)= 1/2+ 3/2= 2 but then you write 2- 1= 1! Which just happens to be the correct answer!

Yes that is my mistake.

Also, how do you know that this triangle is the largest triangle that will fit into that circle? What about the equilateral triangle with vertices equally spaced around the circle?

No, I don't know i just made a guess and draw the picture. Is there a step by step method available to do? Please advise.

Finally, part (b) says "inside that parabola". What parabola?

I did not understand, i think it is printing mistake, but not sure.

However, how did you prove that the triangle with the largest area is the one you drew?

No i have not yet proved, i don't know how to do it. I am just guessing. Please advise.

 

[Dr.Peterson]

I imagine that you are expected to find the area of the triangle (using the usual formula) given that the apex is at a given point (x, y) on the circle, and use differentiation to find the value of x for the maximum area. That way you are not just guessing where it is!

 

[JeffM]

No i have not yet proved, i don't know how to do it. I am just guessing. Please advise.

Well this is much clearer. Dr. Peterson has a suggestion, which is that you are supposed to assume that the apex of the triangle is on the circle. I am not at all sure that that assumption is supported by the text of the problem.

Let's assume that the length of the base is b, which we will consider a constant for now, and the height is h, which for now is the independent variable. Therefore

[MATH]0 < b \le 2 \text { and } 0 < h \le 1.[/MATH]
Do you follow that completely? If not ignore, what follows and let's discuss.

And a for area is the dependent variable.

[MATH]a = 0.5bh \implies \dfrac{da}{dh} = 0.5b.[/MATH]
Do you know where that comes from?

If there is a local maximum, then that derivative will equal 0 there. But this derivative is definitely positive. There is no local maximum. The area simply increases as h increases. But h is bounded. Therefore the area will be maximized when h hits its limit, which is 1.

Now we turn around and treat b as a variable with h = 1. The function describing area is now

[MATH]a = 0.5b * 1 = 0.5b.[/MATH]
What is [MATH]\dfrac{da}{db}?[/MATH]
Where is it zero?

What does the answer to the previous question mean?

So what is the maximum area?