problem of analytical geometry, find area function How this problem is solved?
Determine the "t" function that represents the area of the rectangle inscribed within the following graphs
$$f(x)=x^2-4x$$
and
$$g(x)=\frac{4x-x^2}{2}$$
Edit There was an error in g (x)

 A: This one is a tad bit weird but I'll try my best. To determine the area we only need the height and the width. We can define the area of the nth (we'll label it as x) rectangle by simply defining x as the point at which one vertical side intersects the x-axis.For your tall rectangle it is both ~0.8 and ~3.2 though we only really need to use 1. Right. The vertical height is simple enough. It's merely f(x)-g(x). The width is a tad bit more complex. We can exploit it the symmetry of the curves as they both have the turning point two. So the width is just 2(2-x) [we multiply by two because it's only half the area]. Right then. Now we expand and simplify:
$x^{2}-4x-\frac{4x-x^{2}}{2}=3x^{2}-12x$
$\frac{\left(3x^{2}-12x\right)\left(4-2x\right)}{2}=\frac{-6x^{3}+36x^{2}-48x}{2}=-3x^{3}+18x^{2}-24x$
Now I'm not entirely sure what you wanted t(x) to be but I think it is the above equation. i.e:
$t(x)=-3x^{3}+18x^{2}-24x$
Since it was unclear whether you wanted the largest rectangle or not, see below...
All we do now is find the positive peak of this equation to find the largest area. For cubic formulas that is:
$\frac{\left(-b+\sqrt{b^{2}-3ac}\right)}{3a}$ (or at least that is what I understood).
So the x co-ordinate that represents the point at which a vertical side of the largest possible rectangle intersects with the x-axis is:
$\frac{\left(-\left(18\right)+\sqrt{\left(18\right)^{2}-3\left(-3\right)\left(-24\right)}\right)}{3\left(-3\right)}=2-\left(\frac{2}{\sqrt{3}}\right)$
This is equal to around 0.84.
A: The following diagram may explain some givens:

Blue line is for $f(x)$, the Black line is for $g(x)$.
The following argument should be valid in the interval where g(x)>f(x). That is:
$$x\in [0.845,3.155]$$
The rectangle will have a horizontal side between points $x_1,x_2$ is: $$x_2-x_1$$
From the graph we see that the height is $$g(x_1)-f(x_1)$$
Now we can write the area of the rectangle $A$ as:
$$A=(x_2-x_1)(g(x_1)-f(x_1)) \tag1$$
I guess the important next step is to determine the general equation for $x_2$ given $x_1$.
This is simple knowing that $f(x_1)=f(x_2)$, we can write:
$$x_1^2-4x_1+4=x_2^2-4x_2+4$$
Arrange to get:
$$(x_1^2 - x_2^2)-4(x_2-x_1)=0$$
$$x_2=4-x_1$$
Using this value in (1), results:
$$A=(4-x_1-x_1)((f(x_1)-g(x_1)) $$
That is:
$$A=(4-2x_1)((f(x_1)-g(x_1)) $$
Note, maybe you mean by t function t(x), that represents the area. In this case:
$$t(x)=(4-2x)((f(x)-g(x)) $$
Test:
You can verify the result for example for $x=1$.
A: For convenience, let's shift the functions two units to the left so that the figure is centered at the origin:
$$\begin{align}
f_\star(x) &= f(x+2) = (x+2)^2-4(x+2)= \phantom{-\frac12(}x^2-4 \tag{1}\\
g_\star(x) &= g(x+2) = \phantom{(x+2)^2-4(x+2)} = -\frac12(x^2-4) \tag{2}
\end{align} $$

Notice that $f_\star(x) = -2g_\star(x)$. The negative flips the curve over the $x$-axis, and the $2$ stretches the curve vertically. This tells us that, for whatever (axis-aligned) rectangle we inscribe, the area under the $x$-axis will be exactly twice that above the $x$-axis. Consequently, we need only calculate the area above the axis, and multiply by $3$. Likewise, since the figure is symmetric across the $y$-axis, we need only calculate the area to the right of that axis, and multiply by $2$. 
That is, we need only find the area bounded by a(n axis-aligned) rectangle with one vertex at the origin and one at a point $P=(p,q)$ on the graph of $y=g_\star(x)$, and multiply by $2\cdot 3 =6$. The area of the rectangle is therefore simply
$$A_\star(p) = 6\cdot p q = 6 p\cdot g_\star(p) = -3 p(p^2-4) \tag{$\star$}$$
If the goal were to maximize the area, we could do the standard Calculus things with $A_\star(p)$, which are straightforward and left to the reader, if needed. 
However, the question does not actually mention maximizing; it only asks for the area function (presumably with parameter $t$ that I have inadvertently renamed $p$). This area function is not $A_\star(p)$, because $A_\star(p)$ incorporates shifted coordinates. The target area function would need to be shifted back:

$$A(p) = A_\star(p-2) = -3(p-2)((p-2)^2-4) = -3p(p-2)(p-4) \tag{$\star\star$}$$

(Sanity check: When $p=0$ or $4$, the inscribed rectangle degenerates to a horizontal line segment; when $p=2$, it's a vertical segment. And we have $A(0)=A(2)=A(4)=0$, as expected.) $\square$
