# Optimization for shapes?

The problem is posted here: http://nrich.maths.org/5673

I tried subbing in values to get the biggest area: perimeter ratio possible. I used an arbitrary value, 12, for perimeter. For the triangle, I took out the hypotenuse and so A = 1/2(x)(y) where x+ y = 12. Its area is biggest when x = y = 6 so it has an area of 36. For the rectangle I took out the longer side of the rectangle and so 2x + y = 12. A = x * y. I differentiated and found that its area is biggest when x = 3 and y = 6, giving a area of 18. Now for the circle I was lost on how to approach it since the arc length and area vary in a different manner. I just used a semicircle and guessed that its area would be the largest so I used pi * r = 12. and A = (pi * r^2)/2. r = 12/pi and I subbed that in for area and came up with 22.9183 for area. I have no idea if my logic behind this is correct so I would appreciate any feedback/solutions to this problem. Thanks!!

## Answers and Replies

Simon Bridge
Science Advisor
Homework Helper
The problem is posted here: http://nrich.maths.org/5673

I tried subbing in values to get the biggest area: perimeter ratio possible. I used an arbitrary value, 12, for perimeter.
Because 12 divides by a lot of numbers?

Why not leave it as a variable ... if the required length is p, then A=f(p).
You want to maximize R=A/p = f(p)/p ~ kp since area goes with length-squared.
So actually you are looking for the biggest k.

For the triangle, I took out the hypotenuse and so A = 1/2(x)(y) where x+ y = 12. Its area is biggest when x = y = 6 so it has an area of 36.
case in point:

for x=y=p/2, $$A=\frac{1}{2}\frac{p}{2}\frac{p}{2}=\frac{p^2}{8}$$ ... so k=1/8.

Of course this is for a triangle that has a right-angle in the corner.

For the rectangle I took out the longer side of the rectangle and so 2x + y = 12. A = x * y.
Not bad the short sides are x right?
As drawn, y=2x ... which agrees with your results.

y+2x = 4x =p
y=2x

A=xy = x(2x)=3x^2 3p^2/16 so k=3/16 > 1/8 so the rectangle is better than the triangle.

Now for the circle I was lost on how to approach it since the arc length and area vary in a different manner. I just used a semicircle and guessed that its area would be the largest so I used pi * r = 12. and A = (pi * r^2)/2. r = 12/pi and I subbed that in for area and came up with 22.9183 for area. I have no idea if my logic behind this is correct so I would appreciate any feedback/solutions to this problem. Thanks!!
You want to use the same approach you used before so you can be confident with your results... remove part of the circumference and see which amount removed gives the largest area.

remember: ##C=2\pi R## and ##A=\pi R^2## ... for you ##C=p+x## so x is the amount you have removed. But be careful - can the water fill the entire area A given by that equation?
Does it matter?

But you can cheat - since the other two are the absolute biggest areas for the given length, and the circle is your last shape, you only need to find one circle-shape whose area is bigger than the rectangle one to prove that the circle is the better shape. The question does not ask for the dimensions.

You can just put p=12 if you find it easier to think that way ... the advantage of keeping the variable is that your argument becomes general.

A=xy = x(2x)=3x^2 3p^2/16 so k=3/16 > 1/8 so the rectangle is better than the triangle.
Do you mean A = 2x^2? and thus k = 2/16 = 1/8? so the rectangle and triangle would be the same? and for the circle shape, if I use a semi-circle, then pi*r = P? and thus A = 1/2*pi*r^2 = P^2/(2pi). and k = 1/(2pi) which is greater than 1/8?