# Thread: 80cm of wire cut into two pieces forming a square and a circle

1. ## 80cm of wire cut into two pieces forming a square and a circle

Hello everyone.

Wish i could answer this question on my own,

The question is from october november 2008 AS level maths paper 1 exam paper

Question:

A wire, 80cm long is cut into two pieces. One piece is bent to form a square of side x cm and the other piece is bent to form a circle of radius r cm The total area of the square and circle is A cm^2

(i) show that A=
(pi+4)x^2 - 160x + 1600
-----------------------------
pi

I have no idea how to show that...

then the second question is just as difficult.

(ii) Given that x and r can vary, find the value of x for which A has a stationary value.

Would really love some help

Thanks
Peder

2. Solution:

perimeter of square =4x
perimeter of circle=2 pi r
then 4x+2 pi r=80
r=(40-2x)/pi

Area of square= $\displaystyle x^2$
Area of circle= $\displaystyle pi r^2$

Total area A=$\displaystyle x^2+pir^2$
A=$\displaystyle x^2+pi(40-2x/pi)^2$
A=$\displaystyle x^2+(1600+4x^2-160x)/pi$
A=$\displaystyle (pix^2+1600+4x^2-160x)/pi$
A=$\displaystyle [(pi+4)x^2-160x+1600]/pi$

3. Hello, Peder!

A wire 80cm long is cut into two pieces.
One piece is bent to form a square of side $\displaystyle x$ cm
and the other piece is bent to form a circle of radius $\displaystyle r$ cm.
The total area of the square and circle is $\displaystyle A$ cm³.

(i) Show that: .$\displaystyle A\:=\:\frac{(\pi+4)x^2 - 160x + 1600}{\pi}$

slovakiamath gave an excellent derivation.

(ii) Given that $\displaystyle x$ and $\displaystyle r$ can vary,
find the value of $\displaystyle x$ for which $\displaystyle A$ has a stationary value.

Differentiate $\displaystyle A$, equate to zero, and solve for $\displaystyle x.$

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