A S Level Nephew Problem

[kuiperbelt]

A farmer is to enclose an area of land using a dry stone wall and 200 metres of picket fencing.
The rectangular plot is for home grown produce to meet his ever growing family's needs.
You must show your working and proof. See diagram in the link below:

https://ibb.co/fQXXRtR

[bobbinwales]

Maximum area would be from a square..that's your answer???

[drmorgans]

200 / 3 = 66 2/3 per side

Or did I miss something?

[Prudie]

I'm not going to show the working this time but I would suggest you know
2x+ y=200 (the total length of picket fence)
Area A =x times y
Substitute y=200-2x into the area equation
Then differentiate with respect to x and equate that to zero to find the maximum value of x (from memory differentiate again and if the answer is negative then you have a maximum, which you do in this case)
I won't say what I make x and Y at moment. I might be wrong anyway....

[ClareTG0ld]

I guess the challenge is to *prove* that it's a square. And when you do that, I think you will find that it's not a square.

Let the length of the shorter side be x, and the longer side y, then the area is

A = xy = x(200-2x) = 200x - 2x^2

where I have used x + y + x = 2x + y = 200 from the fixed length of fencing.

A is maximised when dA/dx = 0, or 200 - 4x = 0, or x = 50 -- in which case, y = 100, and A = 5000 m^2.

So, not a square after all! Note that the previous two solutions lead to an area of 4444.4... m^2, which is less.

The key point that was overlooked (by all of us, I was initially expecting it to be a square too) is that the *total* perimeter isn't fixed -- if it had been, then the answer *would* have been a square shape. But it isn't so it's not.

[ClareTG0ld]

My turn to mess up Prudie's withholding of information... :P

[Prudie]

:-)

[drmorgans]

Yep, I missed something. Thanks for the proof.

[bobbinwales]

ok A levels not for me... or at least not doing questions on my tea break after just looking at my phone... shows need to read question properly

[ZebuSanctuary]

Area = x.y

y = 200 - 2x Now substitute for y in Area;

A = x(200 - 2x) Expand brackets A = 200x - 2x²

Differentiating A wrt x

dA/dx = 200 - 4x

At a maxima or minima dA/dx = 0

Therefore 0 = 200 - 4x

Clearly x = 50 metres

Since the fence is 200 metres long, the y dimension = 100 metres

A = x.y = 50.100 = 5000 m²

Proof (Pierre de Fermat?);

Differentiating wrt x the first derivative

dA/dx = 200 - 4x yields

d²A/dx² = -4

Since the second derivative is Negative, it concludes the occurence of x=50 is a maxima.

https://ibb.co/6HwYqjQ

[kuiperbelt]

Some great answers but Zebu's sketch gets BA.

I had not realised 'x' could be 100 metres to make Area = zero
I'm guessing y=0 becaus the 100metres is folded back on itself? LOL

Thanx again everyone!!