Let $y=f(x)$ be a decreasing concave function satisfying $f(a)=0$ for some $a > 0$. Let $S = \{(x,y): 0 \leq x \leq a, 0 \leq y \leq f(x)\}$ be the subset of the positive quadrant under the curve $y = f(x)$.

I'm trying to understand how $S$ can be approximated by rectangles. The smallest possible rectangle $R \supset S$ has vertices $(0,f(0)),(a,f(0)),(a,0),(0,0)$. Its area is $a \cdot f(0)$.

If I take $x = a/2$, then I can create an inscribed rectangle $r$ with vertices $(0,0), (x,0), (x,f(x)),(0,f(x))$. The area of this rectangle is $$Area(r) = x f(x) = a/2 \cdot f(a/2) \geq a/2 \cdot (1/2 \cdot f(0) + 1/2 \cdot f(a)) = a/2 \cdot 1/2 \cdot f(0) = a/4 \cdot f(0) \cdot \geq 1/4 \cdot Area(R)$$ where the inequality follows from the concavity of $f$.

Is 1/4 the optimal ratio for a general concave $f$? If I set $f(x) = \sqrt{1-x^2}$, then I can achieve the ratio 1/2, but I don't know how to get this higher approximation ratio for general $f$.

  • $\begingroup$ Concave or convex? Your title and your question do not match. $\endgroup$ Commented Aug 12, 2021 at 16:15
  • $\begingroup$ The area under a concave function is a convex set. The curve is a convex curve $\endgroup$
    – Asterix
    Commented Aug 12, 2021 at 16:16
  • $\begingroup$ Just try an affine function. $\endgroup$ Commented Aug 12, 2021 at 16:17

1 Answer 1


Let $f(x)=1-x$. Then $f$ is concave and $a=1$. If all rectangles must share axis with the cartesian plane, then the largest inscribed rectangle has an area of $1/4$ (maximize $x(1-x)$ over $[0,1]$). The smallest circumscribed rectangle has an area of $1$, giving a ratio of $1/4$.

Suppose now that there exists decreasing positive concave $f$ with a ratio less than $1/4$. Then, $f(a/2)<f(0)/2$ (else the rectangle you mentioned would have more than a quarter of the area of the circumscribed rectangle), and the function $f$ isn't concave. Therefore, $1/4$ is the optimal (lowest) ratio of the areas of the inscribed rectangle and the circumscribed rectangle.


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