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 ?
 A: 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;
$$x^2 + y^2 = \frac{1}{2}$$
A: 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.
A: 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.
