A 2 = x 2+y 2 ≥ 2xxxxy ≤ a 2/2 (the equal sign is obtained at x=y), so the maximum area is a 2/2. At this time, x = y, that is, when the rectangle is square, its area is the largest.

Solution 2,

See the figure below: let the cylinder volume be v,

V=π*r*r*h

From the cross section in the figure, we can know the satisfaction relation:? r^2+(h/2)^2=R^2？

Involving three times, the construction method adopted:

(a+b+c)/3≥3 radical sign (abc), that is, the arithmetic mean of n positive numbers is not less than its geometric mean (equal sign condition is that the integers used are equal).

R 2 = R 2+(h/2) 2 = R 2/2+R 2/2+H 2/4 ≥ 3 times (cubic root formula (R 2/2 * R 2/2 * H 2/4))? (Take the equal sign when r 2/2 = r 2/2 = h 2/4), at this time, h 2 = 2r 2 is simultaneous with r 2+(h/2) 2 = r 2?

It is obtained that R 2 = 2/3R 2H 2 = 4/3R 2.

According to the inequality, r * r * h ≤ 4 √ 3/9 * r 3 is obtained.

What about when r=√6/3*R? When h=2√3/3*R, the volume can be the maximum value V (max) = π 4 √ 3/9 * r 3.