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= $x^2$
Area of circle= $pi r^2$

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

3. Hello, Peder!

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

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

slovakiamath gave an excellent derivation.

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

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

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

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