# Ratio of inscribed and circumscribed rectangles for a convex set?

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 $$f$$.

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

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, $$f(a/2) (else the rectangle you mentioned would have more than a quarter of the area of the circumscribed rectangle), and the function $$f$$ isn't concave. Therefore, $$1/4$$ is the optimal (lowest) ratio of the areas of the inscribed rectangle and the circumscribed rectangle.