1. Jun 28, 2010

Angela16

The figure shows a triangle ABC where AB=AC, BC=18cm and angle A is a right angle. PQRS is a rectangle with PQ=2x and QR= y cm. Find the area of PQRS and show it in terms of x.
Hence find the maximum area of the rectangle PQRS.

i tried to solve it by myself but i cant do it. Help me,

thanks a lot ^_^

2. Jun 28, 2010

WackStr

Could you explain the problem a little more. How does the triangle relate to the rectangle?

I'm guessing this is a problem in lagrange multipliers, where we need to maximize A = 2xy, with a certain restraint on the variables that needs to yet describe.

3. Jun 28, 2010

Angela16

rectangle located inside of the triangle,
PQRS is given by A=18x-2x square
i missed this part sorry ^_^

i hope that it can help for solve it

4. Jun 28, 2010

WackStr

Firstly note that for any 2x and y, we want the rectangle to have two of it sides running over the base and height of the triangle and one of its corners touching the hypotenuse to maximize the area.

With this observation, we get a relation between 2x and y. (note that i'm putting P over A here):

$$y = 9\sqrt{2}-2x$$

Since the area of the rectangle is $$A = 2xy = 18\sqrt{2}x-4x^2$$, you can use calculus to show that the maximum value of A occurs at $$x = \frac{9\sqrt{2}}{4}$$ and the value of this maximum area is $$A = 40.5$$

As a side-note, notice how this is half the area of the triangle itself. Perhaps you could've concluded this without using calculus as follows:

Add a similar triangle (but flipped) to make a square. Then you'll easily see how maximum area covered will be half the area of the triangle.

I'm mentioning this because I don't know whether you want a calculus based solution or not.

I hope this helps... :)

5. Jun 29, 2010

Angela16

thanks a lot, it was great help ^__^