Optimisation problem (minimum area of a parallelogram inside another parallelogram)

Can anyone help with this question please?

A parallelogram ABCD is given, where AB = 8, BC=4 and BÂC=30º. Let P, Q, R and S be four points, belonging respectively to the sides AB, BC, CD and AD, such that AP = BQ = CR = DS = x. Determine the value of x which gives the smallest area possible of the parallelogram PQRS, specifying what the minimum value of this area is.

The solution should be x=3 and area =7 and the expression for the area of the parallelogram PQRS should be A(x)=x^2 -6x+16...

No procedure is given, so I am a bit stuck

Thank you

Reply 1

mqb2766

Original post by skypestro

Can anyone help with this question please?
A parallelogram ABCD is given, where AB = 8, BC=4 and BÂC=30º. Let P, Q, R and S be four points, belonging respectively to the sides AB, BC, CD and AD, such that AP = BQ = CR = DS = x. Determine the value of x which gives the smallest area possible of the parallelogram PQRS, specifying what the minimum value of this area is.
The solution should be x=3 and area =7 and the expression for the area of the parallelogram PQRS should be A(x)=x^2 -6x+16...
No procedure is given, so I am a bit stuck
Thank you

Have you sketched it and marked on the angles etc?
You could go for calculating the area of PQRS directly or think of it as the area of ABCD and subtract the 4 triangles.

Reply 2

skypestro

Thank you very much for your reply mqb2766

I have tried to do that, but I can't find a way to come up with the required quadratic expression A(x) = x^2-6x+16...

If you ave the patience to follow my steps,you can find them in the attachment. Otherwise, here is an outine of the steps I followed. 1) I found the area of the outer parallelogram (ABCD), 2) then found the sum of the areas of the inner triangles in terms of x, 3) then I subtracted the sum from the area of the outer parallelogram (ABCD) to find an expression for the area of the inner one (PQRS), 4) then I found the derivative of the expression to find the value of x that would give the minimum area, and I did get x=3. However, 5) the minimum area I obtained makes no sense...

Thank you

https://drive.google.com/file/d/1j0spNgwRwDS7j5TiweveBD7u4XRTZIu7/view?usp=drive_link

Reply 3

mqb2766

Original post by skypestro

Thank you very much for your reply mqb2766
I have tried to do that, but I can't find a way to come up with the required quadratic expression A(x) = x^2-6x+16...
If you ave the patience to follow my steps,you can find them in the attachment. Otherwise, here is an outine of the steps I followed. 1) I found the area of the outer parallelogram (ABCD), 2) then found the sum of the areas of the inner triangles in terms of x, 3) then I subtracted the sum from the area of the outer parallelogram (ABCD) to find an expression for the area of the inner one (PQRS), 4) then I found the derivative of the expression to find the value of x that would give the minimum area, and I did get x=3. However, 5) the minimum area I obtained makes no sense...
Thank you
https://drive.google.com/file/d/1j0spNgwRwDS7j5TiweveBD7u4XRTZIu7/view?usp=drive_link

Cant access your drive, can you upload a pic using the camera icon. Note I did it in my head last night and the quadratic/min area you posted is similar but Im not sure its right. Upload what youve tried anyway.

Id guess that completing the square is slightly more elegant/efficient than differentiating and substituting, but both are fine.

(edited 3 weeks ago)

Reply 4

skypestro

Hi,
I've managed to get 7\sqrt{3}, but not 7 as they require...

Anyway, here is my procedure...

Thank you again.

Optimisation 1 copy.jpeg

Reply 5

mqb2766

Original post by skypestro

Hi,
I've managed to get 7\sqrt{3}, but not 7 as they require...
Anyway, here is my procedure...
Thank you again. Optimisation 1 copy.jpeg

Thats what I got and the areas are a sqrt(3) multiple of what you originally posted. One way to check is A(0) should correcpond to the area of ABCD, and the laltter is clearly 16sqrt(3).

You could simplify the working a bit by noting that the parallelograms angles must be 60/120, and then its largely simple trig. Also your sketch is a bit misleading as <ABC must be 60, rather than 120. It doesnt affect the algebra though.

(edited 3 weeks ago)

Reply 6

skypestro

Thanks 👍️

I really appreciate

Reply 7

mqb2766

Original post by skypestro

Thanks 👍️
I really appreciate

Is there any chance you could upload a pic of the original question? If ABC=30, then the ans in the original post would be correct. However you put BAC=30, which gives the ans derived above.