
MHB Apprentice
#1
January 16th, 2020,
08:42
Suppose, a rectangle circumscribes a quadrilateral having length of diagonals p and q, and area A.
What is the maximum area of rectangle that circumscribes the given quadrilateral?
Answer:
How to answer this question using geometry or calculus or by using both techniques.

January 16th, 2020 08:42
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MHB Journeyman
#2
January 17th, 2020,
16:00
Any other information about the "given" quadrilateral other than diagonal lengths, p and q, and area A?

MHB Apprentice
#3
January 17th, 2020,
20:49
If I can see correctly, no more information is needed. Just remember three details:
 Area of a quadrilateral is $ \displaystyle A = \tfrac{1}{2}pq\sin \phi$, where $ \displaystyle p$ and $ \displaystyle q$ are the diagonal lengths and $ \displaystyle \phi$ the angle between the diagonals.
 If you write a variable $ \displaystyle z$ as $ \displaystyle z = a + z  a$, nothing changes, but it might help solving considerably.
 Zeros of the function derivative gives you the critical points inside a closed interval; end points you need to check by substituting.
Hope this helps!

MHB Apprentice
#4
January 18th, 2020,
04:26
Thread Author
Originally Posted by
Theia
If I can see correctly, no more information is needed. Just remember three details:
 Area of a quadrilateral is $ \displaystyle A = \tfrac{1}{2}pq\sin \phi$, where $ \displaystyle p$ and $ \displaystyle q$ are the diagonal lengths and $ \displaystyle \phi$ the angle between the diagonals.
 If you write a variable $ \displaystyle z$ as $ \displaystyle z = a + z  a$, nothing changes, but it might help solving considerably.
 Zeros of the function derivative gives you the critical points inside a closed interval; end points you need to check by substituting.
Hope this helps!
Hello,
I also know the formula of the area of quadrilateral you mentioned, but how to use it here?
Author has given following multiple choices to select:
$$a)\frac{p^2+q^2}{2}$$
$$b)\frac{pq}{2}+A$$
$$c)p^2+q^22A$$
$$d)pq$$
$$e)\frac{p^2+q^2}{4}+A$$
$$f)\frac{p^2+q^2}{3}+3A$$
I don't understand which one to choose as correct answer. Would any member here reply stating the correct answer and reasons thereof. Step by step correct answer is required.
Last edited by Dhamnekar Winod; January 18th, 2020 at 04:47.

MHB Apprentice
#5
January 18th, 2020,
07:22
Originally Posted by
Dhamnekar Winod
I also know the formula of the area of quadrilateral you mentioned, but how to use it here?
Before I provide you the more detailed answer, one last tip:
The area formula gives you the angle $ \displaystyle \phi$, which allows you to write the sides of the rectangle. Draw a good picture, it should help too!

MHB Seeker
#6
January 18th, 2020,
08:06
Since you have given the possible answers, we can use a different approach.
Suppose we pick the square quadrilateral.
That is $p=q$ and $A=pq=p^2=q^2$.
Then the biggest circumscribed rectangle is the square that makes a 45 degree angle with this square quadrilateral.
That is, the maximum rectangle has area $p\sqrt 2\cdot q\sqrt 2= 2pq = 2A$.
So we found a case where the maximum area of the circumscribed rectangle is $2A$.
Unfortunately none of the given possible answers yields $2A$ in this case.
So there is something wrong with the problem statement and the given answers.

MHB Journeyman
#7
January 18th, 2020,
10:52
Originally Posted by
Theia
 Area of a quadrilateral is $ \displaystyle A = \tfrac{1}{2}pq\sin \phi$, where $ \displaystyle p$ and $ \displaystyle q$ are the diagonal lengths and $ \displaystyle \phi$ the angle between the diagonals.
Did I miss where $\phi$ was given?
Originally Posted by
Klaas van Aarsen
Since you have given the possible answers, we can use a different approach.
Suppose we pick the square quadrilateral.
That is $p=q$ and $A=pq=p^2=q^2$.
Then the biggest circumscribed rectangle is the square that makes a 45 degree angle with this square quadrilateral.
That is, the maximum rectangle has area $p\sqrt 2\cdot q\sqrt 2= 2pq = 2A$.
So we found a case where the maximum area of the circumscribed rectangle is $2A$.
Unfortunately none of the given possible answers yields $2A$ in this case.
So there is something wrong with the problem statement and the given answers.
Max area also 2A ... ?

MHB Apprentice
#8
January 18th, 2020,
11:30
Originally Posted by
skeeter
Did I miss where $\phi$ was given?
The area of the quadrilateral is given; as well as the length of both diagonals. Hence we can use the formula of the area of a quadrilateral and solve the angle $ \displaystyle \phi$ in terms of the diagonals and area.

MHB Apprentice
#9
January 18th, 2020,
12:07
Thread Author
Originally Posted by
Theia
The area of the quadrilateral is given; as well as the length of both diagonals. Hence we can use the formula of the area of a quadrilateral and solve the angle $ \displaystyle \phi$ in terms of the diagonals and area.
Hello,
I computed $\sin{\theta}=\frac{2A}{pq}$ . Now what is the next step? Which answer would you choose from the answers a,b,c,d,e,f given by author(#4) as correct answer? and why?

MHB Apprentice
#10
January 18th, 2020,
12:45
You'll need the $ \displaystyle \phi$ later.
Next step: you need to express the sides of the rectangle in terms of the diagonals and some variable you have to introduce for that.
I try to add a picture a bit later. Need to figure out first how it's done...
//To be added: I draw this picture to help my solving process (click to see it fully):
Last edited by Theia; January 18th, 2020 at 13:16.
Reason:
Added picture