# Optimisation problem - circle and square

A piece of wire of length $20$cm is cut into $2$ parts. the first part is bent into a circle of radius $r$ in cm, the second into a square of side length $s$ in cm.

a) write down an expression for the sum of the perimeters of the two shapes in terms of r and s. use this to express $s$ in terms of $r$

I have got $2πr+4s=20$ but don't even know if this is right or not

b) find an expression for $S$, the sum of the areas enclosed by the two shapes in terms of r

c) use differentiation to determine the value of $r$ for which $S$ is a minimum

Really struggling with this as all other examples ask for minimum and maximum areas. I can't even figure out where to start so would appreciate any help!

• Do you know how compute the areas of each shape? – Alex Silva Jan 4 '15 at 18:35

Your answer to a is the first step. Now you should solve it for one of the two variables. $s$ will work better in what follows. For b, what is the area of a square of side $s$? What is the area of a circle of radius $r$? Add them together to get the total area. Now substitute the expression you got in a for $s$ and you have the total area as a function of $r$. The first equation shows you the relation between $s$ and $r$ to use up all the wire. As you increase $s$, you must decrease $r$.
• @skeeto If "1/2$\pi$r" means $\frac12 \pi r$ then your calculations are correct. After differentiating, you had a function of $r$. If you set that function equal to zero, can you solve for $r$? – David K Jan 4 '15 at 19:23
• $\pi$ is just a constant. Your right side for the sum of areas is correct (except for the lack of parentheses-it is not clear that by $1/2\pi r$ you mean $\pi r/2$), but the left side is not $s$ but the total area. When you differentiated you made an error, losing parentheses again. The whole term (5-\pi r/2) should be multiplied by 2 and by the derivative of what is inside the parentheses-you did not multiply the $5$ part. – Ross Millikan Jan 4 '15 at 23:31