# Thread: Problems involving maximization of Area

1. ## Problems involving maximization of Area

Hi MHF!

Now, I'm having a lot of difficulty regarding maximization of area.

The thing I don't know is, how to get the formula of the area to be maximized.

Here are two problems

(MHF is not helping at all, I can't seem to upload anything here!)

This first problem asks to maximize the area of the rectangle, and to give the measure of the maximized triangle.

The other one

The point D is in the minor arc AC.

Can someone help me?

The main problem here is, how to get the area of the thing to be maximized?

Thanks!!

2. You can embed an image hosted somewhere else using the tag [IMG]http://url.com/yourimage.gif[/IMG]

For the first problem, cot(A) = 8/6, so, using trigonometry, you can find sin(A) and express A'A'' through x. Triangle ABC is similar to A'B'C, so B'C = 4x/3, which lets you find A'B'. Now you can express the area of the grey rectangle through x. Note that A'A'' is proportional to 6 - x and A'B' is proportional to x, so the area is proportional to x(6 - x). This means the maximum is achieved when x = 3. To calculate the exact area, you need to find the coefficients of proportionality.

3. Wow! Amazing emakarov!
Thanks for the [IMG] help too, now it is working.

This is really helping me!
Coefficients of proportionality... ok, got it.

I'm getting the hang of geometry now!

So, for the other problem

The point D is in the minor arc AC.

Given that the radius is $\displaystyle \sqrt{7}$
This can be calculated with the Law of Sines.

4. Here we need to maximize the area of ACD since the area of ABC is fixed. Since AC is fixed, the are will be the greatest when the altitude from D to AC is the greatest, i.e., when D is halfway between A and C.

Angle ADC = half of arc ABC = 120, so the area of ADC is 1/2 x AD x DC x sin(120) (cross product of DA and DC). To find AD = DC, you can consider B' halfway between A and C near B. Then B'D is a diameter, so angle B'AD = 90. Angle AB'C = angle ABC = 60, so AB'D = 30, from where AD = 1/2 B'D, i.e., AD equals the radius.

5. Yet again emakarov, thanks for your help. You are correct!

I hope you had fun doing both of them, since they pose some challenge.

Your explanation was great, but I still have some things to I need to clear up.

If I may,

To get the area of ADC you used the formula 1/2 x a x b x sen A?
I didn't understand the cross product explanation very well, can you go into details?
I never understood that formula well.

And one more thing:
Is there some formula to calculate this? What logic did you used here?

One explanation may be that angle B'AD=90 then A must be at the midpoint between B' and D right?
But using the same logic, in a inscribed polygon there can only be a 90 degree angle when the angle is in the midpoint of the diameter, right?

I hope you -or someone else- can answer this.

Thanks again emakarov.

6. To get the area of ADC you used the formula 1/2 x a x b x sen A?

Is there some formula to calculate this? What logic did you used here?

One explanation may be that angle B'AD=90 then A must be at the midpoint between B' and D right?
I am not sure what you mean by "midpoint." To me, a midpoint of a segment is the middle of the segment, and it lies on the segment. Here, A does not lie on B'D. However, since B'D is a diameter (this has to be formally proved, but this is intuitively clear since both B' and D are middles of the two arcs with ends A and C) and since A lies on the circle, angle B'AD = 90 (see inscribed angle theorem in Wikipedia). Angle AB'D = 1/2 angle AB'C = 60, so angle AB'D = 30 and AD = B'D sin(30) = 1/2 B'D.

7. Thanks for the links, I'll give them careful study.

After I studied a little more about inscribed triangles I realised that an inscribed triangle with a right angle doesn't need to be in a midpoint.
What I have meant by midpoint is -the term probably isn't right, is it?- a point in the exact middle of two arcs, in this case two points of the diameter.
But since angle B' is 30, this cannot happen.
I understand your explanation now, thanks for posting again.

emakarov, could you explain the first problem's explanation a little bit further?

For the first problem, cot(A) = 8/6, so, using trigonometry, you can find sin(A) and express A'A'' through x. Triangle ABC is similar to A'B'C, so B'C = 4x/3, which lets you find A'B'. Now you can express the area of the grey rectangle through x. Note that A'A'' is proportional to 6 - x and A'B' is proportional to x, so the area is proportional to x(6 - x). This means the maximum is achieved when x = 3. To calculate the exact area, you need to find the coefficients of proportionality.
So, the first thing you used here was cot? Did you meant tg by that?
This may be common sense but how could you find sin(A)?
$\displaystyle tg (A)=sin(A)/cos(A)$ right?

The other thing is ABC similar to A'B'C
the solution you got here was B'C = 4x/3
This is B'C/A'C=BC/AC
But how come we get such solution?

Next, you used A'A'' is proportional to 6 - x and A'B' is proportional to x
I wouldn't see that as a possibility.
Is this A'A''/6-x and x/A'B'?

So, why did you need the Sin(A) and the measure B'C?
Because, after getting x=3 you can simply use it in a formula and get the exact measures.

And yes, I could be better in Geometry.
But this will help me to clear a lot of things in my mind.
Thanks emakarov!

8. Originally Posted by Zellator
So, the first thing you used here was cot? Did you meant tg by that?
This may be common sense but how could you find sin(A)?
$\displaystyle tg (A)=sin(A)/cos(A)$ right?
By cot(A) I meant cotangent of A, sometimes denoted by ctg(A). You are right, it is the tangent of A that is 8/6. One can find sin(A) through cot(A):

$\displaystyle 1+\cot^2x=1/\sin^2x$

which is obtained from $\displaystyle \sin^2x+\cos^2x=1$ by dividing both sides by $\displaystyle \sin^2x$.

The other thing is ABC similar to A'B'C
the solution you got here was B'C = 4x/3
This is B'C/A'C=BC/AC
But how come we get such solution?
I am not sure what you don't understand. Do you agree that the triangles are similar (A'B' is parallel to AB because A'B'B''A'' is a rectangle)? In similar triangles, corresponding sides have lengths in the same ratio: this is the main property of similarity.

Next, you used A'A'' is proportional to 6 - x and A'B' is proportional to x
I wouldn't see that as a possibility.
A'A'' = sin(A)(6 - x), which means, by definition, that A'A'' is proportional to 6 - x with the coefficient of proportionality sin(A). Similarly, A'B'^2 = (4x/3)^2+x^2=25x^2/9, so A'B'=5x/3, which means, by definition, that A'B' is proportional to x with the coefficient of proportionality 5/3.

So, why did you need the Sin(A) and the measure B'C?
Because, after getting x=3 you can simply use it in a formula and get the exact measures.
What formula?

9. Hi emakarov!
The way you do it is so different from anything I've seen.

Yes I studied some proportionality, as you have recommended me.
I have yet to get the real meaning of the coefficient here.
It will probably work like a ratio, right? I'm not really familiar with it yet.

In my first question, you got the B'C=4x/3
The question is, how come 4 and 3?
This has something to do with the sin(A)?

Calculating, in a wrong way or not, I got sin(A)=sqrt(8/14).
The main doubt here is, since we are using triangle ABC and A'B'C you got the result of 4x/3, but since the sides of ABC are 8 and 6, I don't understand how you can get such solution.
This again has to do something with the coefficient of proportionality right? Since we must multiply them by some number to get such result.

What formula?
Since x=3 you can use the Pythagorean identity and use [4(3)/3]^2+3^2=B'A'^2
This gives B'A'=5 which is the right answer.

Then using the proportionality that I was taught:
(I'm not sure you know this one, I've never researched on it)
$\displaystyle (A'A''/6-X)=(8/10)$
This is leg/hyp=leg/hyp; since they are congruent, this must be right.

A'A'' equals (6-x)4/5
Then applying x=3 you get 12/5
Which is again, the right answer.
So the rectangle is 5 by 12/5, in it's maximum area.

I'm not really familiar with your methods.
But it's really good to see something from another point of view.
I really need to give my best on Geometry here.

10. Originally Posted by Zellator
Yes I studied some proportionality, as you have recommended me.
I have yet to get the real meaning of the coefficient here.
It will probably work like a ratio, right? I'm not really familiar with it yet.
Wikipedia suggest the term "constant of proportionality" rather than "coefficient of proportionality." Don't get too hung up on this; this is not really important. I just meant that $\displaystyle A'B' = c\cdot x$ for some constant c, not, say, $\displaystyle x^2$ or $\displaystyle \sqrt{x}$.

In my first question, you got the B'C=4x/3
The question is, how come 4 and 3?
This has something to do with the sin(A)?
As you wrote above,
This is B'C/A'C=BC/AC
which is the case because triangle ABC is similar to triangle A'B'C. I.e., B'C/x = 8/6. What problem do you see here?

Calculating, in a wrong way or not, I got sin(A)=sqrt(8/14).
Cotangent of A = 6/8 = 3/4. $\displaystyle 1/\sin^2(A) = 1 + \cot^2(A) = 1 + (3/4)^2 = 1 + 9/16 = 25/16$, so sin(A) = 4/5.

Then using the proportionality that I was taught:
(I'm not sure you know this one, I've never researched on it)
$\displaystyle (A'A''/(6-X))=(8/10)$
This is leg/hyp=leg/hyp; since they are similar, this must be right.

A'A'' equals (6-x)4/5
Then applying x=3 you get 12/5
Yes, you don't nees sin(A) to find A'A'' in this way. Good job in noticing the fact that triangles A'A''A and ABC are similar.

11. Thanks again emakarov!
You have been invaluable in helping me here!

Your last post has cleared everything for me!
I made some mistakes because I wasn't in the 'right mood'; it's really not that hard, but I supposed it was very hard before even beginning it.
I just got my TOEFL Test results (103/120) ! That has really lifted my spirit!
I feel I can understand much more easily now; math is related with a good feeling towards it too, right?

Thanks for everything!

12. You are welcome.

Congratulations on TOEFL. When I took it, I think the maximum was still 677 points