# Drawing large rectangle under concave curve

Let $f$ be a continuous concave function on $[0,1]$ with $f(1)=0$ and $f(0)=1$. Does there exist a constant $k$ for which we can always draw a rectangle with area at least $k\cdot \int_0^1f(x)dx$, with sides parallel to the axes, in the area bounded by the two axes and the curve $f$?

If concavity is not required, it is possible to adapt from this example by using the curve $c/x$ to ensure that any rectangle has sufficiently small area. But with concavity, we know that $f$ lies above the line connecting the points $(0,1)$ and $(1,0)$, hence must have area at least $1/2$. If $f$ is exactly that line, then $k=1/2$ exactly. Otherwise, if $f$ is above the line, it looks like the rectangle will even get larger compared to the area under the curve.

• i think it should be $\frac{1}{4}$ since this is the maximum area of a rectangle you can place in the positive quadrant below $\mathrm{conv}((0,1),(1,0))$: the area of the rectangle $[0,\frac{1}{2}]^2$. – Max Jun 20 '16 at 21:59
• @Max But the integral is $1/2$, so the ratio is $1/2$.. – pi66 Jun 20 '16 at 22:02
• I think the area of rectangle is at most $1$, and the integral is as large as possible (consider $f(x)=-ax^2+(a-1)x+1$ with $a>0$, the area is $\frac a6-\frac12$, get larger as $a$ larger), so $k=0$? – user202729 Jun 23 '16 at 7:50
• @user202729: The area of the rectangle isn't bounded by $1$; it can extend beyond $y=1$; it's only parallel to the axes; it needn't use the $y$ axis as one side. – joriki Jun 23 '16 at 8:36
• @joriki I was waiting to see if there would be a better constant, and then I just forgot about it. :( Thanks for your answer! – pi66 Jul 2 '16 at 5:50

Let $t\in[0,1]$ be a value such that $f(x)\le f(t)$ for all $x\in[0,1]$. (Such a value exists since $f$ is unimodal; I don't think we need continuity for this.) The triangle formed by $(0,0)$, $(1,0)$ and $(t,f(t))$ lies under the curve. Its area is $\frac12f(t)$, and it contains an axis-parallel rectangle with half its area, $\frac14f(t)$. The area under the curve is at most $f(t)$. Thus $k=\frac14$ suffices. I'm not sure this is the best possible constant, though.
• The area of that triangle is $f(t)/2$ and the area under the curve is at most $f(t).$ – zhw. Jun 23 '16 at 9:16