maximisation and minimisation (1 Viewer)

maths lover · May 13, 2011

hello,
i was just doing my study when i came across this question which i cant seem to answer.
a string 72cm long is to be cut in two pieces. one piece is used to form a circle and the other a square. what should the perimeter of the square in order to minimize the sum of the two areas.
thx in advance

nightweaver066 · May 13, 2011

72cm. Not sure if my working out is correct.

Amogh · May 13, 2011

SpiralFlex · May 13, 2011

Let the radius of the circle be $x$ and the length of the square be $y$ .

So we can now say the perimeter of the circle and square is $2\pi x+4y = 72$

Now, let's solve for $x$ :

$\therefore x = \frac{72-4y}{2\pi}$

$x$ is the radius of the circle and how do we find the area? Simple.

Area of circle = $(\frac{72-4y}{2\pi})^2 \pi$

Remember we said length of square equals to $y$ ?

So Area of the square is obviously $y^2$

Now the Sum of the areas is the Area of circle + Area of the square.

The Sum of the areas is $(\frac{72-4y}{2\pi})^2 \pi + y^2$

$A = (\frac{72-4y}{2\pi})^2 \pi + y^2$

$= (\frac{5184-576y + 16y^2 }{4\pi})+ \frac{4\pi y^2 }{4\pi}$

$=(\frac{ 4\pi y^2 + 16y^2 -576y+5184 }{4\pi})$

$= (\frac{4(\pi + 4)y^2 -576y+5184}{4\pi})$

$= \frac{4(\pi + 4)y^2}{4\pi}-\frac{576y}{4\pi}+\frac{5184}{4\pi}$

$= \frac{(\pi + 4)}{\pi}y^2 - \frac{144}{\pi}y+ \frac{1296}{\pi}$

Now, remember our Axis of symmetry?

Axis of symmetry $=\frac{144}{\pi} * \frac{\pi}{2(\pi + 4)}$

$= \frac{72}{(\pi + 4)}$

Remember this is our LENGTH of square for when it is minimum. So what is the PERIMETER?

$P = 4 \frac{72}{(\pi + 4)} cm$

$P = \frac{288}{(\pi + 4)} cm$

You got this question off Cambridge!

maths lover · May 13, 2011

yep this came from cambridge, thanks by the way

maximisation and minimisation (1 Viewer)

maths lover

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nightweaver066

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Amogh

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SpiralFlex

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