# Thread: Optimization help needed..

1. ## Optimization help needed..

Help please,
I need to find the max size of these red rectangles in the attached picture.
I started off with a circle of 1ft radius, I found that the large square is 2ft^2
now I dont know how to maximise the area the red squares can be? please help
Thanks

2. Originally Posted by wolfhound
Help please,
I need to find the max size of these red rectangles in the attached picture.
I started off with a circle of 1ft radius, I found that the large square is 2ft^2
now I dont know how to maximise the area the red squares can be? please help
Thanks
In the attachment, the rectangle area is 2(e+c)s
The angle A ranges from 45 to 90 degrees.

From the circle centre $\displaystyle SinA=s+\frac{1}{\sqrt{2}}$

$\displaystyle s=SinA-\frac{1}{\sqrt{2}}$

$\displaystyle TanA=\frac{SinA-\frac{1}{\sqrt{2}}}{c}$

$\displaystyle c=\frac{SinA-\frac{1}{\sqrt{2}}}{TanA}$

$\displaystyle TanA=\frac{\frac{1}{\sqrt{2}}}{e}$

$\displaystyle e=\frac{\frac{1}{\sqrt{2}}}{TanA}$

Rectangle area = 2(e+c)s

Differentiate the product with respect to the angle A
and solve for derivative = 0.

3. Hello
Archie Meade,

Thank you very much for your help and time, I dont fully understand this, is this the easiest way to do it?
Thanks

4. Hi wolfhound.

This involves differentiating a few products of trigonometric terms.
I'll have a think about another way.
If someone doesn't post something simpler, I can elaborate a more thorough explanation later.

5. Hi wolfhound,

here is an easier way to tackle this, but it uses the exact same diagram and the calculations do not differ too much.

The first thing to imagine on the sketch is the height of the green rectangle going to zero. This means the top right hand corner of the rectangle goes down to point "p".

Here, the rectangle area is zero, since it's height is zero, though the base is a maximum.
It corresponds to angle A = 45 degrees (pink line with slope = 1).

Next, consider the point "p" moving to the point "z" with the top right corner of the rectangle attached to it.
At "z", the angle A is 90 degrees, the height is a maximum but the area of the rectangle is again zero, since the base is zero this time.

Somewhere between these extremes, the rectangle will reach a maximum area, as it starts from zero, increases to a maximum and falls to zero again.

To discover this maximum, we use the angle A, pivoted at the circle centre, since the rectangle is in a circle.

There is a little diagram to the left of the circle to help understand the calculations, where you can view the little triangles formed from the angle A.

The area of the rectangle is 2(e+c)s.
The distance from the circle centre to the bottom of the rectangle is $\displaystyle \frac{1}{\sqrt{2}}$
since the circle radius is 1 and $\displaystyle Sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$

We require e, c and s in terms of the angle A.

From the leftmost right-angled triangle....... $\displaystyle TanA=\frac{opp}{adj}=\frac{\frac{1}{\sqrt{2}}}{e}$

Also, from the small top right triangle

$\displaystyle TanA=\frac{s}{c}$

Hence, we can continue with calculations until we have a compact formula for the rectangle area...

$\displaystyle \frac{es}{c}=\frac{1}{\sqrt{2}},\ es=\frac{c}{\sqrt{2}}$

$\displaystyle 2(e+c)s=2es+2cs=\frac{2c}{\sqrt{2}}+2cs=2c(\frac{1 }{\sqrt{2}}+s)$

Area=$\displaystyle 2c\frac{\sqrt{2}s+1}{\sqrt{2}}=\frac{\sqrt{2}\sqrt {2}c}{\sqrt{2}}(\sqrt{2}s+1)=\sqrt{2}c(\sqrt{2}s+1 )$

Substituting for c and s,

$\displaystyle \frac{\sqrt{2}SinA-1}{TanA}[\sqrt{2}SinA-1+1]=\frac{\sqrt{2}SinA-1}{TanA}[\sqrt{2}SinA)$

$\displaystyle =\frac{2Sin^2A-\sqrt{2}SinA}{TanA}=(2Sin^2A-\sqrt{2}SinA)CotA$$\displaystyle =(2Sin^2A-\sqrt{2}SinA)\frac{CosA}{SinA}=(2SinA-\sqrt{2})CosA$

This is a more compact formula for rectangle area.
We need to find the maximum area of the rectangle.
To do that, we differentiate the area formula with respect to A and set the derivative equal to zero.

We use the product rule of differentiation for this.

$\displaystyle \frac{d}{dA}[(2SinA-\sqrt{2})CosA]=CosA(2CosA)+(2SinA-\sqrt{2})(-SinA)$

$\displaystyle 2Cos^2A-2Sin^2A+\sqrt{2}SinA=2(1-Sin^2A)-2Sin^2A+\sqrt{2}SinA$

$\displaystyle \frac{d}{dA}f(A)=0$

$\displaystyle 2-4Sin^2A+\sqrt{2}SinA=0$

$\displaystyle 4Sin^2A-\sqrt{2}SinA-2=0$

This is a quadratic equation of the form $\displaystyle ax^2+bx+c=0$

Hence $\displaystyle SinA=\frac{\sqrt{2}\pm\sqrt{2+32}}{8}=0.905\ and\ -0.552$

As we are looking for an angle between 45 and 90 degrees, we can eliminate the negative result.

$\displaystyle A=Sin^{-1}0.905=1.13 radians$

This angle corresponds to maximum area.

Hence, maximum area is $\displaystyle (2Sin1.13-\sqrt{2})Cos1.13=0.168\ units^2$

6. Here's a couple of graphs to get a visual idea,
solving for the 'angle A' corresponding to maximum rectangle area,
and the graph of rectangle area from A=45 to 90 degrees.

7. Originally Posted by wolfhound
Help please,
I need to find the max size of these red rectangles in the attached picture.
I started off with a circle of 1ft radius, I found that the large square is 2ft^2
now I dont know how to maximise the area the red squares can be? please help
Thanks
Here is a different approach:

The rectangle has the dimensions l and w and the area is $\displaystyle a = l \cdot w$

According to the attached sketch

$\displaystyle l = 2x$

$\displaystyle w = \sqrt{r^2-x^2} - \tfrac12 \cdot \sqrt{2}$

Thus

$\displaystyle a(x)=2x \cdot \left(\sqrt{r^2-x^2} - \tfrac12 \cdot \sqrt{2} \right) = 2x \cdot \sqrt{r^2-x^2} - x \sqrt{2}$

To simplify the following calculations plug in r = 1. Differentiate a(x) wrt x and solve for x

$\displaystyle a'(x) = 0$

You'll get 2 possible solutions and you have to determine which one belongs to the maximum.

8. Earboth's solution is in fact simpler, wolfhound.

9. Originally Posted by Archie Meade
Here's a couple of graphs to get a visual idea,
solving for the 'angle A' corresponding to maximum rectangle area,
and the graph of rectangle area from A=45 to 90 degrees.
Thank you very much for that,its an interesting way of getting it, I will practice it until I can get it.

10. Originally Posted by earboth
Here is a different approach:

The rectangle has the dimensions l and w and the area is $\displaystyle a = l \cdot w$

According to the attached sketch

$\displaystyle l = 2x$

$\displaystyle w = \sqrt{r^2-x^2} - \tfrac12 \cdot \sqrt{2}$

Thus

$\displaystyle a(x)=2x \cdot \left(\sqrt{r^2-x^2} - \tfrac12 \cdot \sqrt{2} \right) = 2x \cdot \sqrt{r^2-x^2} - x \sqrt{2}$

To simplify the following calculations plug in r = 1. Differentiate a(x) wrt x and solve for x

$\displaystyle a'(x) = 0$

You'll get 2 possible solutions and you have to determine which one belongs to the maximum.
Hello earboth,
Thank you very much for your help!

11. Hi wolfhound,

You could try completing Earboth's solution too,
if you have trouble differentiating,
just write up a post to ask for clarity,
it's great practice.

12. Hello , yes I will practice earthbots and yours, I will be back for assistance soon
Thanks

13. hello, I done earboths method this morning and got the max area of one red rectangle as wait....

is this correct

14. You should still be getting about 0.168 square feet.
You calculated x=0.23 gives maximum area.

You should have got $\displaystyle \frac{d}{dx}A=2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}-\sqrt{2}$

Then letting this = 0 gives maximum x=0.424

15. Originally Posted by Archie Meade
You should still be getting about 0.168 square feet.
You calculated x=0.23 gives maximum area.

You should have got $\displaystyle \frac{d}{dx}A=2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}-\sqrt{2}$

Then letting this = 0 gives maximum x=0.424
hello I have done it again,
I have A(-0.5)= -0.159 ft^2 absolute max
its closer but the negatives sign?

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