# Optimization question

A rectangular beam will be cut from a cylindrical log of diameter 1m.

For part a) I have shown that the beam of maximal cross-sectional area is a square.

Then 4 rectangular planks will be cut from the 4 sections of the log that remains after cutting the square beam. Determine the dimensions of the planks that will have maximal cross-sectional area.

So here's what I think: Let x be the width of the rectangular planks and y be the length of the rectangular planks.

Then the Area of one rectangular plank= xy

From part a) The maximal cross-sectional area is 0.5m^2

How do I proceed after this ?

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Don't know if I made a mistake, but I found the maximal cross-sectional area for the first part to be 0.125 $m^2$ – George Tomlinson Oct 17 '13 at 12:00
I found that the sides are (1/2)^1/2 so the Area is 1/2 – Fred Oct 17 '13 at 12:05
maybe there's a mistake in your differentiation? – Fred Oct 17 '13 at 12:05
I didn't use differentiation. It seems to me that if you draw a line from one corner of the maximal square to the other, this line will be the diameter of the circle, which is 1m. This means the distance from a corner of the square to the centre of the square is 0.5m. Then I used Pythagoras to work out the length (calling it 2x) of the sides of the square, as follows: $$0.5^2 = \sqrt {x^2 + x^2} \\ 2x^2 = (\frac {1}{2})^4 \\ x^2 = \frac {1}{32} \\ x = \frac {1}{ \sqrt {32} } = \frac {1}{ 4 \sqrt {2} }. \\$$ The area of the square is $$(2x)^2 = ( \frac {1}{2 \sqrt {2} })^2 = \frac {1}{8}.$$ – George Tomlinson Oct 17 '13 at 12:17
You did a mistake over here it should be 0.5 = sqrt(x^2 + x^2) – Fred Oct 17 '13 at 12:32

You're on the right track. First, draw a diagram indicating the shape of the pieces that remain after the central square is removed. It will have a flat bottom, with the arc of a circle along the top.

Call the location of the edge of a rectangular plank $x$ (so that the plank will have width $2x$). Can you express the height $h$ in terms of $x$? Once you can do that, the area of the plank is

$$A = 2xh(x)$$

and you can choose the $x$ that maximizes that quantity. It will probably be useful to use the equation for the circle;

$$x^2 + y^2 = \frac{1}{2}$$

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I don't understand why the width of the plank is 2x. I thought since the width is x so the A=xh(x). And to how do I find h(x) ? Pythagoras Theorem ?? – Fred Oct 17 '13 at 12:00
I think Chris might mean that you call the origin of the circle $(0,0)$ and then the upper edge of the rectangle is $y$ units higher than the origin, so the length of the rectangle is $2y$. – George Tomlinson Oct 17 '13 at 12:05
I am assuming that the centre of the cylinder is at $(0,0)$ so that one edge of your plank is at $-x$ and the other is at $+x$, for a width of $2x$. Yes, to find $h(x)$ you can use Pythagoras' theorem. – Chris Taylor Oct 17 '13 at 12:06
@Chris Thanks but now I don't seem to be able to relate H and x via Pythagoras' theorem. – Fred Oct 17 '13 at 12:16
See my answer for how to do this @Fred – George Tomlinson Oct 17 '13 at 13:06

Find the turning point of A(x) by calculating when dA/dx = 0. This will be when A is either a local maximum, minimum or inflection point. You can check which kind(if it is not clear for other reasons) by looking at the second derivative.

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Let the origin of the circle be $(0,0)$, the top edge of the rectangle be y units above the origin and the distance from the origin to the side of the rectangle furthest away from the origin be x (noting that the distance from the origin to the near side is half the side of the maximal sqaure $= \frac { \sqrt {2}}{4}$ and so the width of the rectangle (call it w) is $x - \frac { \sqrt {2}}{4})$. You can express y in terms of x using Pythagoras, since the right angled triangle with y and x a sides has a hypotenuse equal to half the diameter of the log.

Now you can express y in terms of w and hence find the formula for the area of the rectangle (call it A) in terms of w, since $A = wh = w(2y)$, where $h$ is the height of the rectangle. Then you can use differentiation to find out whether there is a value of w (and hence y) which maximises A, and if so, which value that is.

I found there are 2 stationary points: one when w is 0, clearly minimising the area, and the other when $w = \frac {-\sqrt{2}}{4}$, which corresponds to $x = 0$, a minimum for the square, so not acceptable. Thus it seems to me there is no maximal rectangle.

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Hmmm.... I actually got the answer. y^2 + (x+1/4√2)^2 = 0.5^2 This is the first equation then I make y subject and substitute it into A(x)=2xy – Fred Oct 17 '13 at 15:03
I'm not seeing how you got that. Is your x the width of the rectangle or the distance from the origin to the far side of the rectangle? In the first case I get $y^2 + (x+\frac {\sqrt {2}}{4})^2 = 0.5^2$, in the second $y^2 + x^2 = 0.5^2$ – George Tomlinson Oct 17 '13 at 15:13
It seems like you have x = the width of the rectangle, in which case it seems to me that $y^2 + (x+\frac {\sqrt {2}}{4})^2 = 0.5^2$.Oh, that is what you have. I thought you were saying $y^2 + (x+\frac {1} {4 \sqrt {2}})^2 = 0.5^2.$ If you mean $y^2 + (x+\frac {\sqrt {2}}{4})^2 = 0.5^2$, then I agree. – George Tomlinson Oct 17 '13 at 15:20
Yes I meant sqrt2/4 not (1/4)(1/sqrt2) sorry I don't know how to type in the mathematical form on computer – Fred Oct 19 '13 at 3:16
You can find out how to type in math form here: math.stackexchange.com/help/notation and here: meta.math.stackexchange.com/questions/5020/… – George Tomlinson Oct 19 '13 at 11:37