# Maximum area of rectangle which circumscribes the given quadrilateral.

#### Dhamnekar Winod

##### Active member
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?

How to answer this question using geometry or calculus or by using both techniques.

#### skeeter

##### Well-known member
MHB Math Helper
Any other information about the "given" quadrilateral other than diagonal lengths, p and q, and area A?

#### Theia

##### Well-known member
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!

#### Dhamnekar Winod

##### Active member
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:

$$a)\frac{p^2+q^2}{2}$$
$$b)\frac{pq}{2}+A$$
$$c)p^2+q^2-2A$$
$$d)pq$$
$$e)\frac{p^2+q^2}{4}+A$$
$$f)\frac{p^2+q^2}{3}+3A$$

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.

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#### Theia

##### Well-known member
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!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

#### skeeter

##### Well-known member
MHB Math Helper
- 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?

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 ... ?

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#### Theia

##### Well-known member
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.

#### Dhamnekar Winod

##### Active member
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?

#### Theia

##### Well-known member
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):

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#### Theia

##### Well-known member
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.
I checked this and according to my solution everything is right. My solution agrees with case b).

#### Dhamnekar Winod

##### Active member
I checked this and according to my solution everything is right. My solution agrees with case b).
Hello,
So as per your work, the maximum area of rectangle circumscribing the given quadrilateral having diagonals p and q, and area A, is $pq\cos{\alpha}*\cos{\beta}$. That means $pq\cos{\alpha}*\cos{\beta}=\frac{pq(1+\sin{\theta})}{2}\tag{1}$ because you have selected answer b).

How would you prove equation (1)?

#### Theia

##### Well-known member
Just plug in the expression for $$\displaystyle \beta$$, find $$\displaystyle \alpha$$, which gives the maximum area and see where you'll land. I think it's enough for proof, if done properly.

#### Dhamnekar Winod

##### Active member
Just plug in the expression for $$\displaystyle \beta$$, find $$\displaystyle \alpha$$, which gives the maximum area and see where you'll land. I think it's enough for proof, if done properly.
Hello,
$pq\cos{\alpha}*\cos{(\frac{\pi}{2}-\theta-\alpha)}=pq\cos{\alpha}*\sin{(\theta+\alpha)}\tag{2}$

Now $\sin{(a+b)}=\sin{a}\cos{b}+\cos{a}\sin{b}$. After plugging this in (2), how to get R.H.S. of (1) in #12?

#### Theia

##### Well-known member
Let's define the circumscribing quadrilateral has area $$\displaystyle A_c$$, and now we have obtained:

$$\displaystyle A_c(\alpha) = pq\cos \alpha \cdot \sin (\phi - \alpha)$$.

Now it's time to use the 3rd tip in my first post: the derivative. We must find the angle $$\displaystyle \alpha$$ such that $$\displaystyle A_c(\alpha)$$ has a maximum. So:

What is $$\displaystyle \frac{\mathrm{d}A_c(\alpha)}{\mathrm{d}\alpha}$$?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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):
Here you go.

\begin{tikzpicture}

\newcommand\anglephi{atan(3/2)}
\newcommand\anglealpha{(90-atan(3/2))/2}
\coordinate[label=below right:O] (O) at (0,0);
\coordinate[label=left:A] (A) at (-3,0);
\coordinate[label=above:B] (B) at (2,3);
\coordinate[label=right:C] (C) at (6,0);
\coordinate[label=below] (D) at (-3,-4.5);
\coordinate[label=left:A'] (A') at ({180-\anglealpha}:{3*cos(\anglealpha)});
\coordinate[label=above:B'] (B') at ({\anglephi+\anglealpha}:{sqrt(2^2+3^2)*cos(\anglealpha)});
\coordinate[label=right:C'] (C') at ({-\anglealpha}:{6*cos(\anglealpha)});
\coordinate[label=below'] (D') at ({180+\anglephi+\anglealpha}:{sqrt(3^2+4.5^2)*cos(\anglealpha)});

\draw[thick] (O) node at +({\anglephi/2}:1) {$\phi$} +(0.6,0) arc (0:\anglephi:0.6);
\draw[thick] (O) node at +({\anglephi + \anglealpha/2}:1.4) {$\alpha$} +({\anglephi}:1) arc (\anglephi:\anglephi+\anglealpha:1);
\draw[thick] (O) node at +({180 - \anglealpha/2}:1.4) {$\beta$} +({180 - \anglealpha}:1) arc (180-\anglealpha:180:1);
\draw[rotate={-90-\anglealpha}] (A') rectangle +(0.3,0.3);
\draw[rotate={-90-\anglealpha}] (B') rectangle +(0.3,0.3);
\draw[rotate={90-\anglealpha}] (C') rectangle +(0.3,0.3);
\draw[rotate={90-\anglealpha}] (D') rectangle +(0.3,0.3);
\draw[rotate={-90-\anglealpha}] (A') ++(B') rectangle +(0.3,0.3);
\draw[rotate={90-\anglealpha}] (C') ++(D') rectangle +(0.3,0.3);

\draw (A) -- (C);
\draw (B) -- (D);
\draw (A') -- (C');
\draw (B') -- (D');
\draw[blue, ultra thick] (A) -- (B) -- (C) -- (D) -- cycle;
\draw[red, ultra thick] (A') -- +(B') -- (B') -- +(C') -- (C') -- +(D') -- (D') -- +(A') -- cycle;

\end{tikzpicture}

#### Dhamnekar Winod

##### Active member
Let's define the circumscribing quadrilateral has area $$\displaystyle A_c$$, and now we have obtained:

$$\displaystyle A_c(\alpha) = pq\cos \alpha \cdot \sin (\phi - \alpha)$$.

Now it's time to use the 3rd tip in my first post: the derivative. We must find the angle $$\displaystyle \alpha$$ such that $$\displaystyle A_c(\alpha)$$ has a maximum. So:

What is $$\displaystyle \frac{\mathrm{d}A_c(\alpha)}{\mathrm{d}\alpha}$$?

Hello,
Assuming $A_c(\alpha)=pq*\cos{\alpha}\sin{(\theta+\alpha)}$ and after computing its derivative and doing some algebra i got the answer $\frac{pq}{2}*(1+\sin{\theta})$. But one other math expert has given the answer to this question in another way. I didn't understand some part of that answer. I shall reproduce it tomorrow for your review so that you will explain its difficult part to me.

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello,
Assuming $A_c(\alpha)=pq*\cos{\alpha}\sin{(\theta+\alpha)}$ and after computing its derivative and doing some algebra i got the answer $\frac{pq}{2}*(1+\sin{\theta})$. But one other math expert has given the answer to this question in another way. I didn't understand some part of that answer. I shall reproduce it tomorrow for your review so that you will explain its difficult part to me.
We can rewrite it as:
$$A_c=\frac{pq}{2}\cdot(1+\sin{\phi}) = \frac{pq}{2}+\frac 12 pq \sin\phi=\frac{pq}{2}+A$$
which is indeed consistent with the given solution b).

#### Dhamnekar Winod

##### Active member
We can rewrite it as:$$A_c=\frac{pq}{2}\cdot(1+\sin{\phi}) = \frac{pq}{2}+\frac 12 pq \sin\phi=\frac{pq}{2}+A$$which is indeed consistent with the given solution b).
Hello, Different way to answer this question. Suppose that the diagonals of the quadrilateral make an acute angle $\alpha$ with each other, so that $A=\frac12pq\sin{\alpha}$ Consider a rectangle that circumscribes the quadrilateral. If the diagonal of length p makes an angle of $\theta$ with the side of rectangle that it touches, then the rectangle must have height $p\sin{\theta}$ and width $q\cos{(\theta-\alpha)}$. So that the rectangle has area$$\Delta=pq\sin{\theta}\cos{(\theta-\alpha)}=\frac12 pq[\sin{(2\theta-\alpha)+\sin{\alpha})}]=\frac12 pq\sin{(2\theta-\alpha)}+A.$$so that $\Delta \leq \boxed{\frac12 pq+A}$ with the maximum achieved when $\theta=\frac12\alpha+\frac{\pi}{4}$Now, in this answer how the width of rectangle is computed. Secondly what is the symbol $\Delta?$ Is it representing the area of the rectangle?Thirdly, how area of rectangle= $\frac12 pq\sin{(2\theta-\alpha)}+A.$How the statement $''\Delta \leq\boxed{\frac12pq +A}$ with the maximum achieved when $\theta=\frac{\alpha}{2}+\frac{\pi}{4}"$ is made? My answer for maximum $\theta$ is $\frac{\pi}{4}-\frac{\theta}{2}$Would any member explain answers to these queries to me? I also want to know whether the same method has been used to answer this question as previously we used.

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#### Theia

##### Well-known member
I'm on go now, so I can miss something, but at least I can see these two:

Secondly what is the symbol $\Delta?$ Is it representing the area of the rectangle?
Yes

So that the rectangle has area

$$\Delta=pq\sin{\theta}\cos{(\theta-\alpha)}=\cdots$$

Thirdly, how area of rectangle= $\frac12 pq\sin{(2\theta-\alpha)}+A.$

How the statement $''\Delta \leq\boxed{\frac12pq +A}$ with the maximum achieved when $\theta=\frac{\alpha}{2}+\frac{\pi}{4}"$ is made?
It's done by looking at the expression of the area:

$$\displaystyle \Delta = \frac12 pq\sin{(2\theta-\alpha)}+A$$

using the fact $$\displaystyle |\sin x| \le 1.$$

My answer for maximum $\theta$ is $\frac{\pi}{4}-\frac{\theta}{2}$
You have a mistake here: if you have a critical point for $$\displaystyle \theta$$, you most likely don't want to define it implicitly i.e. using the $$\displaystyle \theta$$ in nonsolved form, as explicit form i.e. $$\displaystyle \theta = \mathrm{something}$$. Can you check this, what was meant?

#### Dhamnekar Winod

##### Active member
I'm on go now, so I can miss something, but at least I can see these two:

Yes

It's done by looking at the expression of the area:

$$\displaystyle \Delta = \frac12 pq\sin{(2\theta-\alpha)}+A$$

using the fact $$\displaystyle |\sin x| \le 1.$$

You have a mistake here: if you have a critical point for $$\displaystyle \theta$$, you most likely don't want to define it implicitly i.e. using the $$\displaystyle \theta$$ in nonsolved form, as explicit form i.e. $$\displaystyle \theta = \mathrm{something}$$. Can you check this, what was meant?

Hello,
Sorry. I made mistake in my last post. Please read "My answer for maximum $\alpha=\frac{\pi}{4}-\frac{\theta}{2}"\tag{1}$

My $\frac{dA_c(\alpha)}{d\alpha}=pq\cos{(2\alpha+\theta)}$. Now to maximise the area of rectangle $2\alpha+\theta=\frac{\pi}{2}$ Hence the result (1).

Now our area of rectanle=$pq\cos{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\sin{\left(\frac{\pi}{4}+\frac{\theta}{2}\right)}=\frac{pq}{2}+A$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello,
Different way to answer this question.

Suppose that the diagonals of the quadrilateral make an acute angle $\alpha$ with each other, so that $A=\frac12pq\sin{\alpha}$
Consider a rectangle that circumscribes the quadrilateral. If the diagonal of length p makes an angle of $\theta$ with the side of rectangle that it touches, then the rectangle must have height $p\sin{\theta}$ and width $q\cos{(\theta-\alpha)}$. So that the rectangle has area

$$\Delta=pq\sin{\theta}\cos{(\theta-\alpha)}=\frac12 pq[\sin{(2\theta-\alpha)+\sin{\theta})}]=\frac12 pq\sin{(2\theta-\alpha)}+A.$$

so that $\Delta \leq \boxed{\frac12 pq+A}$ with the maximum achieved when $\theta=\frac12\alpha+\frac{\pi}{4}$

Now, in this answer how the width of rectangle is computed?
The angles are labeled differently than in the picture we posted a couple of posts ago.
In particular we have:
$$\theta_{new}=\frac\pi 2 - \beta_{old} \\ \alpha_{new} = \phi_{old} \\$$
In that picture the sides of the circumscribed rectangle have the same length as A'C' and B'D'.
We have:
$$A'C' = AC \cos\beta_{old} = p\cos(\frac\pi 2 - \theta_{new}) = p\sin\theta_{new} \\ B'D' = BC\cos\alpha_{old} = q\cos(\frac\pi 2-\phi_{old}-\beta_{old}) = q\cos(\theta_{new}-\alpha_{new})$$

I also want to know whether the same method has been used to answer this question as previously we used.
It is the same method.
That is, we find an expression for the area of a circumscribed rectangle as function of an angle parameter.
And we find its extremum, which can be done either by taking the derivative and setting it to zero, or by observing that the sine function has a maximum of $1$ at $\frac\pi 2$.