First, work out the perimiter of the rectangle. The length a has length x, the length b has length (10-x). So the perimiter P is given by:

So

, no matter how the rectangle is defined (so long as x>0, y>0.Now, if you know that the biggest rectangle enclosed by a set perimiter is in fact a square... you can easily work out your answer, because for a square,

so and so forth.]]>So we have a rectangle which lies entirely inside the triangle formed by the origin and the two points where the line y = 10 - x crosses the co-ordinate axes. Obviously there are an infinity of such rectangles possible. We want to find the one with the maximum area.

So, let's choose a width for the general rectangle and call it w. Obviously, 0 <= w <= 10. 10 because this is where the line y = 10 - x crosses the x axis.

Now, the height of this retangle must be 10 - w (this is because it has a vertex on the line y = 10 - x).

So, in terms of w, the area of the rectangle is

Now we have to maximise this function. So we differentiate it w.r.t. w which gives:

and set this equation to zero to obtain:

Well at w = 5 we have y = 10 - 5 = 5 therfore the area of this rectangle is base*height = 5 * 5 = 25

And this is a maximum not a minimum as

which is negative everywhere.

Mitch.

]]>In general, when you see something like "find the largest" or "find the leas/smallest" etc, you should imedeately think of calculus - most of these problems are solved using differentiation.

The general way to solve these problems is:

1) Find an equation/equations that describe the problem at hand.

2) Make whichever quantity that you have to maximise/minimise the subject of the equation/equations (in this case, that means you want an equation that looks like *Area = (something)*

3) Differentiate said equation with respect to a "relevant" variable in order to find the maxima/minima of your equation from (2). This variable will be what you're able to change in order to maximise/minimise your answer. In this case, you can change the area of the rectangle by moving it along the y = 10 - x line.

This might seem a bit complicated/abstract for the moment, but have a look at the diagram:

As you can see, there is one degree of freedom in this picture - if you place the corner of the rectangle at a point on the x-axis, it has to lie at a certain place on the y-axis (since it has to lie on y = 10 - x), and this fixes what the area of the rectangle is. Now for the actual maths - first off, finding an equation for the area of the rectangle.

The area will be

. No, a is the point and b will be . Since , we conclude that:Now we've done both (1) and (2) in the list above - we're looking to maximise the value of

, given that we can choose .The next step is to differentiate the above equation for

and find the maximum/minimum values.I'll leave the rest to you for some practice - give it a try and if you need more help, feel free to post again and ask.

]]>Word Problem wrote:

A rectangle has one vertex on the line

y= 10 -x.x> 0, another at the origin, one on the positivex-axis, and one on the positivey-axis. Find the largest areaAthat can be enclosed by the rectangle.

Any help would be very much appreciated!

]]>