Author 
Message 
lucky_luke science forum addict
Joined: 11 Oct 2005
Posts: 51

Posted: Fri Apr 14, 2006 11:05 pm Post subject:
Find max area of rectangle inscribed in triangle



hi
I don't know how to even begin.
I have to find maximum area of rectangle inscribed inside of rectangle with height H and hypotenuse c.
In right triangle with hypotenuse c = 8 and angle = 30 is inscribed rectangle with maximum possible area.
Find sides a and b of triangle, if two sides of rectangle
lie on a and b
thank you 

Back to top 


Lynn Kurtz science forum Guru
Joined: 02 May 2005
Posts: 603

Posted: Sat Apr 15, 2006 12:13 am Post subject:
Re: Find max area of rectangle inscribed in triangle



On Fri, 14 Apr 2006 19:05:01 EDT, lucky_luke <seveda1@yahoo.com>
wrote:
Quote:  hi
I don't know how to even begin.

The way to begin, if you expect to get help, is to state the problem
exactly as it is given to you instead of adding your confusion to the
statement of the problem.
Lynn
Quote:  I have to find maximum area of rectangle inscribed inside of rectangle with height H and hypotenuse c.
In right triangle with hypotenuse c = 8 and angle = 30 is inscribed rectangle with maximum possible area.
Find sides a and b of triangle, if two sides of rectangle
lie on a and b
thank you 


Back to top 


Jasen Betts science forum Guru Wannabe
Joined: 31 Jul 2005
Posts: 176

Posted: Sat Apr 15, 2006 6:46 am Post subject:
Re: Find max area of rectangle inscribed in triangle



On 20060414, lucky_luke <seveda1@yahoo.com> wrote:
Quote:  hi
I don't know how to even begin.

Quote:  I have to find maximum area of rectangle inscribed inside of rectangle with height H and hypotenuse c.

i'll assume the second "rectangle" is a triangle.
Half the area of the triangle. Interestingly the converse is true too.
Bye.
Jasen 

Back to top 


G.E. Ivey science forum Guru
Joined: 29 Apr 2005
Posts: 308

Posted: Sat Apr 15, 2006 12:57 pm Post subject:
Re: Find max area of rectangle inscribed in triangle



Quote:  hi
I don't know how to even begin.
I have to find maximum area of rectangle inscribed
inside of rectangle with height H and hypotenuse c.
In right triangle with hypotenuse c = 8 and angle =
30 is inscribed rectangle with maximum possible area.
Find sides a and b of triangle, if two sides of
rectangle
lie on a and b
thank you

?? You posted several different problems. I assume you mean: given a right triangle find the area of the largest rectangle that can be inscribed in the right triangle with the right angle of the triangle as one corner of the rectangle.
"Find sides a and b of triangle, if two sides of
rectangle lie on a and b" makes no sense at all. I THINK you are saying you have a right triangle with hypotenuse c = 8 and angle = 30 degrees.
Then the first thing you need to do (the rectangle is irrelevant at this point) is find the lengths of sides a and b: imagine the triangle reflected about the side adjacent to the 30 degree angle to make a triangle with angle 60 degrees. It's easy to see that that is an equilateral triangle with side of length 8. Side a (opposite the 30 degree angle) of the right triangle is half the third side of the equilateral triangle and so has length 4. By the Pythagorean theorem, side b, has length sqrt(64 16)= sqrt(4= sqrt(16*3)= 4sqrt(3).
Now, for the rectangle. Here's how I would do it: Set up a coordinate system with x axis along a and y axis along b. One corner of the rectangle is at (0,0), one corner is on the xaxis, one corner is on the yaxis and one corner is on the hypotenuse. The hypotenuse is a line passing through (4, 0) and (0, 4sqrt(3)) and so x/4+ y/(4sqrt(3))= 1. Solving for y: y= sqrt(3)x+ 4sqrt(3). That is, if the the vertex on the xaxis is at (x,0) then the vertex on the hypotenuse is at (x, 4sqrt(3)sqrt(3)x) and so the vertex on the yaxis is at (0, 4sqrt(3) sqrt(3)x). The area of the rectangle is the product of those lengths: A= 4sqrt(3)x sqrt(3)x^2= (sqrt(3)(x^2 4x).
One method of finding maximum area is to take the derivative of that but for a quadratic, we can just complete the square:
A= (sqrt(3))(x^2 4x+ 4 4)= (sqrt(3))((x 2)^24).
(x2)^2 is never negative so (x2)^2 4 is larger than 4 as long as x is not 2 and has minimum value (4) at x= 2. That is, A is less than 4sqrt(3) as long as x is not 2 and has maximum value 4sqrt(3) when x= 2.
That is interesting the triangle has base of length 4 and height of length 4sqrt(3) and so area (1/2)(4)(4sqrt(3)= 8sqrt(3). As Jasen Betts said, the area of the largest rectangle inscribed in this way is half the area of the triangle itself. 

Back to top 


Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Sat Apr 15, 2006 5:45 pm Post subject:
Re: Find max area of rectangle inscribed in triangle



In article
<30982153.1145105880281.JavaMail.jakarta@nitrogen.mathforum.org>,
"G.E. Ivey" <george.ivey@gallaudet.edu> wrote:
Quote:  hi
I don't know how to even begin.
I have to find maximum area of rectangle inscribed
inside of rectangle with height H and hypotenuse c.
In right triangle with hypotenuse c = 8 and angle =
30 is inscribed rectangle with maximum possible area.
Find sides a and b of triangle, if two sides of
rectangle
lie on a and b
thank you
?? You posted several different problems. I assume you mean: given a
right triangle find the area of the largest rectangle that can be
inscribed in the right triangle with the right angle of the triangle as
one corner of the rectangle.
"Find sides a and b of triangle, if two sides of
rectangle lie on a and b" makes no sense at all. I THINK you are saying you
have a right triangle with hypotenuse c = 8 and angle = 30 degrees.
Then the first thing you need to do (the rectangle is irrelevant at this
point) is find the lengths of sides a and b: imagine the triangle reflected
about the side adjacent to the 30 degree angle to make a triangle with
angle 60 degrees. It's easy to see that that is an equilateral triangle
with side of length 8. Side a (opposite the 30 degree angle) of the right
triangle is half the third side of the equilateral triangle and so has
length 4. By the Pythagorean theorem, side b, has length sqrt(64 16)=
sqrt(4= sqrt(16*3)= 4sqrt(3).
Now, for the rectangle. Here's how I would do it: Set up a coordinate
system with x axis along a and y axis along b. One corner of the rectangle
is at (0,0), one corner is on the xaxis, one corner is on the yaxis and
one corner is on the hypotenuse. The hypotenuse is a line passing through
(4, 0) and (0, 4sqrt(3)) and so x/4+ y/(4sqrt(3))= 1. Solving for y: y=
sqrt(3)x+ 4sqrt(3). That is, if the the vertex on the xaxis is at (x,0)
then the vertex on the hypotenuse is at (x, 4sqrt(3)sqrt(3)x) and so the
vertex on the yaxis is at (0, 4sqrt(3) sqrt(3)x). The area of the
rectangle is the product of those lengths: A= 4sqrt(3)x sqrt(3)x^2=
(sqrt(3)(x^2 4x).
One method of finding maximum area is to take the derivative of that but
for a quadratic, we can just complete the square:
A= (sqrt(3))(x^2 4x+ 4 4)= (sqrt(3))((x 2)^24).
(x2)^2 is never negative so (x2)^2 4 is larger than 4 as long as x is
not 2 and has minimum value (4) at x= 2. That is, A is less than 4sqrt(3)
as long as x is not 2 and has maximum value 4sqrt(3) when x= 2.
That is interesting the triangle has base of length 4 and height of length
4sqrt(3) and so area (1/2)(4)(4sqrt(3)= 8sqrt(3). As Jasen Betts said, the
area of the largest rectangle inscribed in this way is half the area of the
triangle itself.

This maximum of "half the area" is true for an arbitrary triangle.
One such rectangle of maximal area may always be formed with one side of
the rectangle along a longest side of the triangle and the opposite side
of the rectangle having vertices at midpoints of the other two sides of
the triangle. 

Back to top 


lucky_luke science forum addict
Joined: 11 Oct 2005
Posts: 51

Posted: Sat Apr 15, 2006 6:24 pm Post subject:
Re: Find max area of rectangle inscribed in triangle



I really did screw up my post and I'm sorry for that.
Here is original question reposted and "repaired"
Quote:  I have to find maximum area of rectangle inscribed
inside of triangle with height H and hypotenuse c.

How would I solve first this,since in that example
vertices of rectangle may very well be located only
on two sides of triangle( if triangle has sides a, b
and c, then two vertices could be located on c,
and third vertex on either a or b,and 4. vertex would
be located somewhere inside a triangle not touching
any of the three sides )?Of course, this would happened
only if one of the angles inside triangle would bigger
that Pi/2. 

Back to top 


Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Sat Apr 15, 2006 9:32 pm Post subject:
Re: Find max area of rectangle inscribed in triangle



In article
<7219404.1145125528144.JavaMail.jakarta@nitrogen.mathforum.org>,
lucky_luke <seveda1@yahoo.com> wrote:
Quote:  I really did screw up my post and I'm sorry for that.
Here is original question reposted and "repaired"
I have to find maximum area of rectangle inscribed
inside of triangle with height H and hypotenuse c.
How would I solve first this,since in that example
vertices of rectangle may very well be located only
on two sides of triangle( if triangle has sides a, b
and c, then two vertices could be located on c,
and third vertex on either a or b,and 4. vertex would
be located somewhere inside a triangle not touching
any of the three sides )?Of course, this would happened
only if one of the angles inside triangle would bigger
that Pi/2.

If the given triangle is to have a hypotenuse, it must be a right
triangle, and one can then choose arbitrarily to have one side of a
maximal rectangle along any one side of the triangle. And if one does
not choose to have the rectangle along the hypoteneuse, two sides of the
maximal rectangle will have to lie along the shorter two sides of the
triangle.
It is only in triangles with one angle greater than a right angle that
the maximal rectangle is limited to lying along only one side, the side
opposite the greatest angle. 

Back to top 


lucky_luke science forum addict
Joined: 11 Oct 2005
Posts: 51

Posted: Sun Apr 16, 2006 10:09 pm Post subject:
Re: Find max area of rectangle inscribed in triangle



Quote:  I really did screw up my post and I'm sorry for that.
Here is original question reposted and "repaired"
I have to find maximum area of rectangle inscribed
inside of triangle with height H and hypotenuse c.
How would I solve first this,since in that example
vertices of rectangle may very well be located only
on two sides of triangle( if triangle has sides a, b
and c, then two vertices could be located on c,
and third vertex on either a or b,and 4. vertex would
be located somewhere inside a triangle not touching
any of the three sides )?Of course, this would happened
only if one of the angles inside triangle would bigger
that Pi/2.
If the given triangle is to have a hypotenuse,
it must be a right triangle, and one can then
choose arbitrarily to have one side of a
maximal rectangle along any one side of the
triangle. And if one does not choose to have the
rectangle along the hypoteneuse, two sides of the
maximal rectangle will have to lie along the
shorter two sides of the triangle.
It is only in triangles with one angle greater
than a right angle that the maximal rectangle
is limited to lying along only one side, the
side opposite the greatest angle.

It was bad choice of words on my part. I shouldn't
use the word hypotenuse since problem didn't
specify it. I don't think we are dealing with right
triangle. Any idea how I would solve it? 

Back to top 


Virgil Hancher science forum beginner
Joined: 16 Apr 2006
Posts: 5

Posted: Mon Apr 17, 2006 12:03 am Post subject:
Re: Find max area of rectangle inscribed in triangle



In article
<28027513.1145225386814.JavaMail.jakarta@nitrogen.mathforum.org>,
lucky_luke <seveda1@yahoo.com> wrote:
Quote:  I really did screw up my post and I'm sorry for that.
Here is original question reposted and "repaired"
I have to find maximum area of rectangle inscribed
inside of triangle with height H and hypotenuse c.
How would I solve first this,since in that example
vertices of rectangle may very well be located only
on two sides of triangle( if triangle has sides a, b
and c, then two vertices could be located on c,
and third vertex on either a or b,and 4. vertex would
be located somewhere inside a triangle not touching
any of the three sides )?Of course, this would happened
only if one of the angles inside triangle would bigger
that Pi/2.
If the given triangle is to have a hypotenuse,
it must be a right triangle, and one can then
choose arbitrarily to have one side of a
maximal rectangle along any one side of the
triangle. And if one does not choose to have the
rectangle along the hypoteneuse, two sides of the
maximal rectangle will have to lie along the
shorter two sides of the triangle.
It is only in triangles with one angle greater
than a right angle that the maximal rectangle
is limited to lying along only one side, the
side opposite the greatest angle.
It was bad choice of words on my part. I shouldn't
use the word hypotenuse since problem didn't
specify it. I don't think we are dealing with right
triangle. Any idea how I would solve it?

Consider putting one side of your rectangle along a longest side of the
triangle with its other vertices on the other two sides and use
properties of similar triangles. 

Back to top 


lucky_luke science forum addict
Joined: 11 Oct 2005
Posts: 51

Posted: Mon Apr 17, 2006 9:45 pm Post subject:
Re: Find max area of rectangle inscribed in triangle



Can someone show me? 

Back to top 


Jasen Betts science forum Guru Wannabe
Joined: 31 Jul 2005
Posts: 176

Posted: Tue Apr 18, 2006 8:39 pm Post subject:
Re: Find max area of rectangle inscribed in triangle



On 20060416, lucky_luke <seveda1@yahoo.com> wrote:
Quote:  It was bad choice of words on my part. I shouldn't
use the word hypotenuse since problem didn't
specify it. I don't think we are dealing with right
triangle. Any idea how I would solve it?

you can divide any triangle into two right triangles and the maximal
rectangles for both will gave the same side length on thee shared part of
the triangle.
the answer's the same; the area of the rectangle will be half that of the
triangle.

Bye.
Jasen 

Back to top 


Google


Back to top 



The time now is Mon Dec 05, 2016 12:28 am  All times are GMT

