Answer:
Explanation:
If r and h radius and of inscribed cylinder then from similyarity of triangles(2- h)/2 = 2r/15, from here h = 2 - 4r/15.
Volume V = πr2h = πr2(2 - 4r/15)= π(2r2 - 4r3/15). Let find derivative and set it to 0:
V' = π(4r - 12r2/15) = 0; we get equation 4r - 4r2/5 = 0, r = 0 and r = 5.
To check if we get maximum we find V'' = π(4 - 8r/5) and at r = 5 V'' = π(4 -8) = -4π < 0 max
Answer: demantions of inscribed cylinder r = 5 and h = 2 - 4·5/15 = 2 - 4/3 = 2/3