Let $f$ be a continous function with $f: I[0,1] \rightarrow \mathbb{R}$ with $f(x)>0$ for all $x\in I[0,1]$. For every Point x we can draw an rectangle parallel to the axis with $(0,0),(x,0),(x, f(x)), (0,f(x))$. Show that there exists a choice for x so that the rectangle has a maximum surface area.
We can determine the surface area with F = a * b, so in this case we have F = d(0-x) * (0-f(x)). But I really don't know how I can get a suitable x.
The area of the rectangle = base * height = $xf(x)$ Draw a diagram and look at it to clarify.
So we need to maximize the product xf(x).
We know that the function y = x is continuous everywhere, and we are given that f is continuous on the closed bounded interval [0,1].
So the product [xf(x)] is continuous on the closed bounded interval [0,1].
It is a standard theorem that a continuous function on a closed bounded region has a minimum and maximum at some point in that region (including the boundary.)