# Area Problem.

## Main Question or Discussion Point

You have a linear function $$f(x)=-\frac{2}{3}x+4$$ where $$0\leq x\leq 6$$ and $$0\leq y\leq 4$$. What is the maximum area of a rectangle that has one side on the line of the function?

I know how to optimize this, I am just having trouble finding the equation for area of such a rectangle in terms of the function.

I have included an image of what I am talking about. Thanks a lot.

Prematurely posted... Fixing.

uart
What you need to do is to get express all the pertinent quantities in terms of a single parameter that can be adjusted. For example consider the starting point of writing equation of the line containing the parallel side as (a - 2/3 x).

In terms of parameter "a" the other quanitities you're interested in are :

y intercept of parallel side, yi = a

x intercept of parallel side, xi = 1.5a,

Length of parallel side, L = a sqrt(3.25)

Length of perpendicular side Lp = 3(4-a) / sqrt(13)

Area, A = 3/2 * a (4-a)

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I don't really get where you got the length of the perpendicular side, but then again, I'm rushing to get to class... I had another way to solve the problem, but it was taking forever to write out that silly latex code stuff.

http://students.washington.edu/marusich/trianglemax.jpg [Broken]

$$\mbox{Note that the area of the rectangle is zero when } x = 0 \mbox{ and when } x = 6 \mbox{. Also, just by looking, we see that the area ought to increase, then}$$
$$\mbox{eventually hit a maximum and decrease, on the } x \mbox{-interval } [0, 6] \mbox{. This sort of thing is useful to observe before getting started, because it either}$$
$$\mbox{helps confirm what we calculate or makes us stop and think when we see something strange (like if we calculated that one of the endpoints}$$
$$\mbox{were the maximum). Anyway...}$$

$$\mbox{A} = ab$$

$$\cos\theta = \frac{x}{a}$$

$$a = \frac{x}{\cos\theta}$$

$$\sin\theta = \frac{b}{6-x}$$

$$b = (6-x)\sin\theta$$

$$\mbox{A} = ab = \frac{x}{\cos\theta}(6-x)\sin\theta$$

$$= x(6-x)\frac{\sin\theta}{\cos\theta} = x(6-x)\tan\theta$$

$$\mbox{but}$$

$$\tan\theta = \frac{4}{6} = \frac{2}{3}$$

$$\mbox{so we just have:}$$

$$\mbox{A}(x) = \frac{2}{3}x(6-x) = 4x-\frac{2}{3}x^2$$

$$\mbox{A'}(x) = 4-\frac{4}{3}x$$

$$\mbox{A}(x) \mbox{ has a max or min either at the endpoints of the } x \mbox{-interval } [0, 6] \mbox{ or when A'}(x) = 0 \mbox{ (or at any other critical point, but since this function is just}$$
$$\mbox{a parabola, it has no asymptotes, kinks, jumps, holes or anything like that). The max should definitely not occur at the endpoints in this}$$
$$\mbox{case (remember our initial observation), or we've done something wrong.}$$

$$\mbox{A'}(x) = 4-\frac{4}{3}x = 0$$

$$\frac{4}{3}x = 4$$

$$x = 3$$

$$\mbox{if }x > 3 \mbox{ then A'}(x) < 0$$
$$\mbox{verify: A'}(10000000000) = 4 - \frac{4}{3}(10000000000)<0$$

$$\mbox{if }x < 3 \mbox{ then A'}(x) > 0$$
$$\mbox{verify: A'}(0) = 4-\frac{4}{3}(0) = 4 > 0$$

http://students.washington.edu/marusich/derivative.jpg [Broken]

$$\mbox{So A}(x) \mbox{ has a global maximum at } x = 3 \mbox{. The endpoints of our particular interval } [0, 6] \mbox{ will not be maxima because A}(x) \mbox{ is increasing}$$
$$\mbox{for all } x < 3 \mbox{ and decreasing for all } x > 3 \mbox{, which means that A}(3) > \mbox{A}(x) \mbox{ for all }x \neq 3 \mbox{. You could verify this by simply comparing}$$
$$\mbox{A}(0) \mbox{ and A}(6) \mbox{ to A}(3) \mbox{.}$$

$$\mbox{So the maximum area is:}$$

$$\mbox{ A}(3) = 4(3)-\frac{2}{3}(3)^2 = 6$$

$$\mbox{ (Still another reason this result makes sense is because the point }(3, 6) \mbox{ lies in quadrant I, which makes sense because A}(x)\mbox{ was an inverted}$$
$$\mbox{parabola with its maximum obviously somewhere in quandrant I.) }$$

Converted to latex. It's the first time I've used it. Hurray for easier-to-read math! Oh, and looking back, I guess you only wanted the relation for the rectanglular area...Oh well. Here's the whole thing. :P

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uart
Night Owl said:
I don't really get where you got the length of the perpendicular side ...
From similar triangles. I'm too lazy to draw a diagram so just refer to Alex's original thumbnail and look at the small right-angle triangle in the upper left hand coner of the figure. The hypotemus of this triangle is (4-a) and one of the other sides is the perpendicular distance. Further this small triangle is similar to the largest triangle in the figure which gives the ratio of these two sides as 3/sqrt(13).

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I see how you got it, now. For future reference, it's usually easier to follow if you show your steps and not just the end result...even if it seems like a simple thing to you.

Anyway, your way also works, and it gives the same result for the area. It's always nice to see another way to do a problem.

uart
Night Owl said:
For future reference, it's usually easier to follow if you show your steps and not just the end result...even if it seems like a simple thing to you.
You know that questions like this one are quite often someones homework or assignment. It's ok to give them some pointers on the route to follow without giving them every detail.

There is a easier way to solve this problem

My english is poor,but I guess you can understand my idea easily.
Firstly,we let g(x)=-(2/3)x+h be our one line of rectangular
when g(x)=0,x=(3/2)h;
when x=0,g(x)=h;
so the two other line of rectangular are p1(x)=(3/2)x+h and p2(x)=(3/2)x-(9/4)h;
let (3/2)x+h=(-2/3)x+4 and (3/2)x-(9/4)h=(-2/3)x+4,you will get the information of other two vertexs,so all four vertexs are only concerned with "h".you can draw a fuction of area S(h) and find its maximum area easily.

I guess I must learn to say something mathmatical in english. :surprised