Maximize the area of a triangle

  • Thread starter Thread starter youngstudent16
  • Start date Start date
  • Tags Tags
    Area Triangle
Click For Summary

Homework Help Overview

The problem involves maximizing the area of a triangle formed by a point on a semicircle and its projections onto the diameter. The semicircle has a radius of 1, and the triangle's vertices are defined by points on the semicircle and the diameter.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the coordinates of points on the semicircle and the area of the triangle. There are attempts to derive the area formula and its derivative. Some participants question the clarity of the notation used, particularly regarding the definitions of variables and the presentation of equations.

Discussion Status

There is ongoing clarification regarding the mathematical expressions and definitions used in the problem. Some participants have provided constructive feedback on how to present the equations more clearly, while others are still working to define their variables and understand the implications of their calculations.

Contextual Notes

Participants note the importance of adhering to homework posting guidelines and the need for clearer definitions of terms such as x, y, and A. There is also mention of the challenges faced when using mobile devices for mathematical notation.

youngstudent16
Messages
59
Reaction score
1
http://imgur.com/Q5gjaSG

Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.

What I'm thinking
y=sqrt(r^2-x^2) From the formula of a circle x^2+y^2=r^2
A=1/2(x+1)y The area of a triangle
A'=(1-x-2x^2)/2sqrt(1-x^2) Substitution from the formula of a circle and take the derivative
A'=0 if x=1/2 Set the derivative = to 0 to maximize it
y=sqrt(3)/2 Solve for y
A=1/2((1/2)+1)sqrt(3)/2= 3sqrt(3)/8 Plug in all the pieces into the original equation of the area of a triangle.

A=Area A'=Derivative of area
x+1 = base of the triangle
y=The heightThese are still new to me let me know how far off I am. I think I have the equation wrong.

(sorry posted this in wrong area first)
 
Last edited:
Physics news on Phys.org
youngstudent16 said:
http://imgur.com/Q5gjaSG

Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.

What I'm thinking y=sqrt(r2-x2) A=1/2(x+1)y A'=(1-x-2x2)/2sqrt(1-x2) A'=0 if x=1/2 y=sqrt(3)/2 A=1/2((1/2)+1)sqrt(3)/2= 3sqrt(3)/8
The above is very difficult to read or make sense of. What are x and y?
Is r2 supposed to mean r2? If so, you should at least write r^2 to indicate that 2 is an exponent.
youngstudent16 said:
These are still new to me let me know how far off I am.

(sorry posted this in wrong area first)

Homework problems should be posted in the Homework & Coursework section, using the homework template.
 
The working may be OK (depending on what you mean by the various symbols), but is spoiled by a careless presentation.

First: what is 'x', and what is its range? (Don't make the reader guess---just tell him/her what you mean).
Second: when you write r2 and x2, do you mean ##r^2## and ##x^2##? If so, write them as r^2 and x^2 in plain text (which is what you would also need to do for numerous software packages if you were using a computer to calculate things).
Third: use parentheses around leading fractions like your '1/2'; in other words, write (1/2)(x+1) y ---- or better, use (1/2)*(x+1)*y in plain text. Parentheses help to avoid confusion between 1/2ab (which can be interpreted as ##\frac{1}{2ab}## and (1/2) ab (which, unambiguously, means ##\frac{1}{2} ab##).
Finally----and maybe most importantly--- break up your sequence of equations into separate lines, or at least, make it clear what goes with what. The way you wrote it looks like
y = \sqrt{r2-x2}A=1/2(x+1)y A' = (1x - 2x2)/2 \sqrt{1-x2} A'=0
instead of the intended (I hope) meaning
y = \sqrt{r^2 - x^2} = \sqrt{1-x^2} \;(\text{since} \: r = 1).
The area is
<br /> A = (1/2)(x+1) A, \;\text{so that the derivative is} \; A&#039; = (x - 2 x^2)/(2 \sqrt{1-x^2}).
The maximal area occurs when ##A' = 0##, or ##x = 1/2##, giving maximal area ##A = 3 \sqrt{3}/8##.

You don't need to use LaTeX (as I have done), but you do need to put in occasional clarifying words/sentences so that others can follow what you are trying to do. Even the simple strategy of putting different things on different lines will help, so you could write
y=sqrt(r2-x2)
A=1/2(x+1)y
A'=(1-x-2x2)/2sqrt(1-x2)
A'=0 if x=1/2, y=sqrt(3)/2, A=1/2((1/2)+1)sqrt(3)/2= 3sqrt(3)/8.

That could get you higher marks, too.
 
  • Like
Likes   Reactions: youngstudent16
I fixed it not sure what the template is but I gave the question and my attempt, What else should I conclude?

^ thank you you broke down what I was trying to say. I'm not aware of latex I will use it in the future I"m on a phone so not sure I can use it though. The x was just the extra distance to get the base of the triangle on top of the radius. That is the part I think I'm messing up with also.
 
Last edited:
youngstudent16 said:
I fixed it not sure what the template is but I gave the question and my attempt, What else should I conclude?

BTW: You don't need quite so many parentheses, so you don't need to say sqrt(r^(2)-x^(2)); just plain sqrt(r^2 - x^2) will do, because the parsing rules for mathematical expressions will take care of the "precedence". You still need to develop the habit of defining your terms: saying what are x, y and A for example. (In the case of x, it is particularly important, because two different people doing this problem might well use two different definitions of x!)
 
  • Like
Likes   Reactions: youngstudent16
Thanks for the feedback I think I tried to define everything a little better in the original post.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K