Help with Yr 11 Ext Maths :jedi: (1 Viewer)

[vaaan]

Heeellooo

So I'm having trouble with a few questions on maximisation and minimisation for school.

Keep in mind you can't use calculus to solve these cause we haven't learnt them at school yet. I think you have to get some general equation and find the vertex and hence the max or min D:

1. Prove that the rectangle of greatest area that can be inscribed in a circle is a square. [HINT: Recall that the maximum of A occurs when the maximum of A^2 occurs.]

2. OAB is a triangle in which OA is perpendicular to OB. OA and OB have lengths of 60cm and 80cm respectively. A rectangle inscribed inside the triangle so that one of its sides lies along the base OA of the triangle. By using similar triangles find the size of the rectangle of maximum area that may be inscribed in the triangle.

3. A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the area of the triangle.

Thanks

 

[Gussy Booo]

:| . What topic is this?

 

[vaaan]

'Problems on Maximisation and Minimisation' according to my textbook D: it's the cambridge one btw. 298-299

 

[Gussy Booo]

This is the Quadratic Function......

Omgsh, my stupid MATHS IN FOCUS BOOK DOESNT COVER THIS !!!

 

[Mountain.Dew]

With Q1, you won't get a quadratic function. You will get something hideous YUCK. you MUST use diff, it can't be done any other way.
WithQ2, you will get a quadratic, so you can use the vertex method. No need for dif (although it is way easier to dif)
With Q3, same with Q2. You will get a quadratic, use vertex method.

Enjoy!

 

[Drongoski]

Let circle have radius 'r'. Let 'x' be the width of any inscribed rectangle. Now this rectangle is symmetric about the centre of the circle with diagonal = 2r. By Pythagoras thm, the height of the rectangle = sqrt(4r² - x²). .: area of rectangle A = x . sqrt(4r² - x²). Therefore A² = x² . (4r² - x²). Now A achieves its maximum when A² achieves its maximum. Therefore finding for what values of 'x' A is max is the same as for A².

Now A² is a quadratic in x² [just like the quadratic in m: m(4r^2 -m) = -m² + 4r² m ]and takes the value 0 @ x² = 0 and 4r². Since coeff of (x²)² of this quadratic is negative,(concave downwards parabola) we have a maximum value for A² at the mid-point of the 2 zeroes of this quadratic = (0 + 4r²)/2 = 2r²; i.e. when x² = 2r² or equivalently for x = sqrt(2) r. The corresp height now becomes sqrt(4r² - 2r²) = sqrt(2) r which is equal to the width 'x' for maximum A. Therefore the rectangle must be a square.

 

[bored.of.u]

just cause u havnt be taught calculus doesnt mean u cant use it. using calculus would make these q's simple.