I have little to no idea how to do this problem...
A rectangle has one vertex on the line y = 10 - x. x > 0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area A that can be enclosed by the rectangle.
Any help would be very much appreciated!
Last edited by Saitenji (2006-09-21 14:57:36)
(edit - just noticed you said "precalculus" in your title! If you don't know what differentiation is, skip this post and see my next one. Sorry!)
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 forand 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.
Last edited by Dross (2006-09-21 21:15:49)
Bad speling makes me [sic]
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.
Last edited by gnitsuk (2006-09-21 20:51:28)
Okay - I just noticed your title said "precalculus"! In which case you would go about things a whole different way:
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.
Bad speling makes me [sic]