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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!

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  • $\begingroup$ Do you know how compute the areas of each shape? $\endgroup$ – Alex Silva Jan 4 '15 at 18:35
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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 $r$.

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  • $\begingroup$ I then get s=5-1/2πr so that should answer part a. i then get s=πr^2+(5-1/2πr)^2 for part be which i understand i must differentiate for part c. for this i got 2πr-5π+1/2π^2r. I think this is correct and I'm sure the next step is to find stationary points but i don't know how to from here $\endgroup$ – skeeto Jan 4 '15 at 18:48
  • $\begingroup$ @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 $r$? $\endgroup$ – David K Jan 4 '15 at 19:23
  • $\begingroup$ yes that's what it should have shown thanks. How would i go about differentiating it? I'm unsure because i've never dealt with an equation with the inclusion of π. $\endgroup$ – skeeto Jan 4 '15 at 20:44
  • $\begingroup$ $\pi$ 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. $\endgroup$ – Ross Millikan Jan 4 '15 at 23:31

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