Maximize the area of a triangle

In summary: I also tried to clean it up a little bit.In summary, the goal of this problem is to maximize the area of the triangle BDC, where B and C are points on a semicircle with radius 1 and D is the projection of C onto AB. Using the formula for a circle and taking the derivative, we can find that the maximum area occurs when x=1/2 and y=sqrt(3)/2, giving a maximal area of 3sqrt(3)/8. The use of parentheses and defining terms such as x and y is important in presenting a clear and organized solution.
  • #1
youngstudent16
59
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
  • #2
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.
 
  • #3
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
[tex] y = \sqrt{r2-x2}A=1/2(x+1)y A' = (1x - 2x2)/2 \sqrt{1-x2} A'=0 [/tex]
instead of the intended (I hope) meaning
[tex] y = \sqrt{r^2 - x^2} = \sqrt{1-x^2} \;(\text{since} \: r = 1).[/tex]
The area is
[tex]
A = (1/2)(x+1) A, \;\text{so that the derivative is} \; A' = (x - 2 x^2)/(2 \sqrt{1-x^2}).[/tex]
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 youngstudent16
  • #4
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:
  • #5
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 youngstudent16
  • #6
Thanks for the feedback I think I tried to define everything a little better in the original post.
 

FAQ: Maximize the area of a triangle

What is the formula for finding the area of a triangle?

The formula for finding the area of a triangle is 1/2 * base * height.

How do you maximize the area of a triangle?

To maximize the area of a triangle, you need to use the formula A = 1/2 * base * height, where base and height are the variables. You can increase the area by increasing the base and height, or by keeping the base and height equal.

What is the relationship between the base and height of a triangle when maximizing the area?

The base and height of a triangle are directly proportional when maximizing the area. This means that when one increases, the other also increases in order to maximize the area.

How can I apply the concept of maximizing the area of a triangle in real-life situations?

The concept of maximizing the area of a triangle can be applied in various real-life situations, such as when designing a garden or building a roof. In both cases, you would want to maximize the area to make the most out of the available space.

Is there a limit to how much you can maximize the area of a triangle?

Yes, there is a limit to how much you can maximize the area of a triangle. The area of a triangle cannot be greater than the area of a rectangle with the same base and height. This is because a rectangle has the largest area among all quadrilaterals with the same perimeter.

Back
Top