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Re: 2012 HSC MX2 Marathon Close. Geometrically, f(x,y)=4xy is the area of a rectangle inscribed inside the given ellipse. Thus the result found geometrically is area of the largest possible rectangle that can be inscribed in the ellipse.

Re: 2012 HSC MX2 Marathon A bit of several variables calculus... (i) Use the method of Lagrange multipliers, or otherwise, to find the maximum value of f(x,y)=4xy on the ellipse x^2 + 4y^2=4. (ii) Give a geometrical interpretation of the result in (i).

Re: 2012 HSC MX2 Marathon What maths is that taught in? Haven't come across it yet...

Who is your lecturer?

In 2U students will probably find it easier to use the perpendicular distance formula. That is, Find the centre of the circle and the radius (by completing the square) and (i) The perpendicular distance from the centre of the circle to the line y=mx should be less than the radius of the...

Re: 2012 HSC MX2 Marathon Yeah I think I realised during uni maths that it was just integrating starting from the "other end"...

Re: 2012 HSC MX2 Marathon Trebla likes his GP type of questions :P

Re: 2012 HSC MX2 Marathon Hahahahah novice error by Trebla!

Most of the questions are pretty much the same with Volumes...

Re: 2012 HSC MX1 Marathon I think you might be able to complete the square on x+x^(-1), but I haven't tried that yet...

Well, you can use cramer's rule. But sometimes there will be infinite solutions (in the case where you have, say, 3 variables but only 2 equations). The matrix method I'm talking about is Gaussian elimination and putting the matrix in row echelon form. In that form you can see immediately if...

Well, you can use cramer's rule. But sometimes there will be infinite solutions (in the case where you have, say, 3 variables but only 2 equations). The matrix method I'm talking about is Gaussian elimination and putting the matrix in row echelon form. In that form you can see immediately if...

:o Jetblack... No longer mod, and you were banned before? :O

This shouldn't really be asked in the HSC. It's more of a linear algebra topic. And if you do it by substitution you always have to check your answers. The matrix method is much nicer and you can tell straight away if there are no solutions to the system of linear equations.

Re: 2012 HSC MX2 Marathon Pro (Y).

Re: 2012 HSC MX1 Marathon Yes, but you can assign a cauchy principal value to the integral since it doesn't converge absolutely and since the integrand is an odd function.

Where did you get the q's from? They looks familiar...

Re: 2012 HSC MX2 Marathon From memory, I think there's an easier way to do this question. Ie. using z z (bar) = |z|^2

Re: 2012 HSC MX2 Marathon I asked that question in 2009, and you answered it oO.

Re: 2012 HSC MX1 Marathon Well, one could still say that \text{p.v. } \int_{-1}^1 \frac{1}{x} \ dx = 0 p.v. means the cauchy principle value.