Hi,

How can I find the largest possible perimeter for a rectangle that can be drawn inside the ellipse
(x/a)² + (y/b)² = 1.

It's only a 2 mark question, but I just can't see how to do it at the moment.

Thanks.

I get 4(a² +b² )^½ - not sure if that's right. Seems to require quite a bit of thinking compared to ordinary 2 mark questions.

Kerwin

I don't actually understand how you got the 4(a² +b² )^1/2 .

Could you please explain?

Kerwin

Perhaps this explanation will seem to make more sense. First, are you ok with what we are trying to maximize is y+x? (Our answer will be 4 times this though). We can use only positive values for y since this is what we are finding. We therefore want to maximize x+b(1-x² /a² )^1/2 . We can solve this by differentiating and setting to zero, then find x, find y, and add them together, then multiply by 4, and you should be able to come up with the answer.

Hope this helps,

Brad

Though, I should mention, Kerwin's method is perfectly fine and a good deal more elegant than this method.

A more elegant method involves the QM-AM inequality.....

What's that method Kerwin? QM-AM: is that something to do with the arithmetic and q***(something) means?

((1-l)x²+ly²)^1/2 ³ (1-l)x+ly

for all 0 £ l £ 1. Using this inequality, we get

x+y £ (a2+b2)[(1-b2/(a2+b2))(a-2x)2+(b2/(a2+b2))(b-2y)2 ]1/2 = (a2+b2)[(x2 a-2+y2 b-2)/(a2+b2)]1/2 = (a2+b2)1/2

Kerwin

What is Cauchy-Schwarz? I've seen it mentioned a couple of times on here, but never asked about it.

Regards,
Olof

Kerwin

This is basically an extension of the fact that the magnitude of a cosine of a real angle is less than or equal to 1.

If:

x = (x₁ ,x₂ )
y = (y₁ ,y₂ )

Then (x₁ y₁ + x₂ y₂ )² = (x .y )² = |x |² |y |² cos² (theta) < = |x |² |y |² = (x₁ ² + x₂ ² )(y₁ ² + y₂ ² )

where theta is the angle between x and y . This demonstrates the inequality for n = 2 and also makes it clear why they need to be parallel for equality. You can justify it in a similar way for N-D though you need to be careful about how you define angle. Alternatively, you can prove the inequality by expanding it out and collecting terms and using the fact that squares are positive.

Here is a previous thread where C-S was discussed about halfway down the page.

Kerwin

Cheers guys.

Olof