# Maximize the area of a triangle

1. Jul 18, 2015

### youngstudent16

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 height

These 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: Jul 18, 2015
2. Jul 18, 2015

### Staff: Mentor

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.
Homework problems should be posted in the Homework & Coursework section, using the homework template.

3. Jul 18, 2015

### Ray Vickson

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
$$A = (1/2)(x+1) A, \;\text{so that the derivative is} \; A' = (x - 2 x^2)/(2 \sqrt{1-x^2}).$$
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.

4. Jul 18, 2015

### 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: Jul 18, 2015
5. Jul 18, 2015

### Ray Vickson

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!)

6. Jul 18, 2015

### youngstudent16

Thanks for the feedback I think I tried to define everything a little better in the original post.