Max and Min Problem (1 Viewer)

swifty13 · Mar 5, 2012

Hi guys if anyone could help me with this question that'd be great.

A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced.

The two formulas that I got were A = xy and x^2+y^2=144, but I had trouble differentiating

SpiralFlex · Mar 5, 2012

First of all, always draw a pretty diagram. *Pending*

$x^2+y^2=144$

$y=\sqrt{144-x^2}$

Take the positive solution because we can't have a negative measurement in the HSC course.

$A=x\sqrt{144-x^2}$

$\frac{dA}{dx}=1*\sqrt{144-x^2}+x * \frac{1}{2}*\frac{-2x}{\sqrt{144-x^2}}$

$=\sqrt{144-x^2}-\frac{x^2}{\sqrt{144-x^2}}$

For maximum or minimum, $\frac{dA}{dx}=0$

$x=\sqrt{72}$

Substitute that back into the equation,

$y=\sqrt{72}$

Therefore maximum area is 72 cm squared. (Of course you need to test for max/min)

This is of course is a square.

With further work, you will find that the largest rectangle with maximum area inscribed in a circle is in fact a square.

Carrotsticks · Mar 5, 2012

If you do the question logically, the largest rectangle that can be produced is a square.

In fact, the largest n-sided polygon inscribed within a circle of any radius is the regular n-sided polygon concentric to the circle in which it was inscribed. Bonus points if you can prove this.

We are given that the radius of the circle is 6cm but we realise that this is the same as the diagonal of the square. Using Pythagoras' Theorem, we see that the length of the square is the sum of the squares of the radius. Using a bit more algebra, we find that the area of the square is 2 x 6^2 = 72 centimetres squared.

However, I presume that you want the rote-learning + 'methodical' way that the BOS tries to encourage (unfortunately). Below is a proof that the largest possible area is indeed the square.

$\\ xy=A \Rightarrow y = \frac{A}{x}\\\\ x^{2} + y^{2} = 144 \\\\ $Substitute into each other:$ \\\\ x^{2} + { \left (\frac{A}{x} \right )}^{2} = 144 \\\\ x^{2} + \frac{A^{2}}{x^{2}}} = 144 \\\\ x^{2} + A^{2} x^{-2} = 144 \\\\ $Differentiate with respect to x:$ \\\\ 2x - 2A^{2} x^{-3} = 0 \\\\ 2x - \frac{2A^{2}}{x^{3}} = 0 \\\\ $Multiply both sides by $ x^{3} $ and then divide by 2$: \\\\ x^{4} - A^{2} = 0 \\\\ x^{4} = A^{2} \\\\ x^{2} = A \\\\ $But we remember that $ A=xy $ so we use this:$ \\\\ x^{2} = xy \\\\ \therefore x=y \\\\$

$\\ $And this proves that maximum area (minimum area obviously occurs when the rectangle is the degenerate case of a single line ie: area of 0) occurs when x =y, which implies that the rectangle is a square. Then proceed to use the geometrical techniques I described above to proceed to find that area.$$

Carrotsticks · Mar 5, 2012

Ohh dear I miread the question... the RADIUS was 6 not the area of the circle.

Allow me to correct my answer accordingly.

Max and Min Problem (1 Viewer)

swifty13

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SpiralFlex

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Carrotsticks

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Carrotsticks

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