Tiêu đề Apr 2025 PB1 - MSTC Kippap Solutions.pdf ✅

From the figure, R1 = R2 = a, Using Pythagorean theorem, ( H 2 ) 2 + a 2 = 102 a 2 = 100 − H 2 4 From the volume of the frustum, V = πH 6 (3a 2 + 3a 2 + H 2 ) = πH 6 (6a 2 +H 2 ) 1396.263 = πH 6 [6 (100 − H 2 4 )+ H 2 ] H = 4. 521 m Alternatively, the top and bottom portions form spherical caps, Vc = πh 2 3 (3R − h) 1396.263 = πh 2 3 [3(10) − h] h = 7.739 m Finally, H = 2(10) − 2(7.739) = 4. 522 m. Problem 4. Optimization Tourists look at a 2 m x 2 m painting from a stand 4 m away. If the eye level of the tourists is 1.5 m above the floor, how high should the bottom of the painting be placed for best possible viewing? a. 4.62 m c. 3.62 m b. 3.12 m d. 4.12 m Solution: The distance for maximum viewing angle is given by x = √h1h2 where h1 is the distance of the top of the object from the eye level and h2 is the distance of the bottom of the object from the eye level. From the problem, htop = hbottom + 2, h2 = hbottom − 1.5, h1 = htop −1.5 = hbottom + 0.5. 4 = √(hbottom +0.5)(hbottom −1.5) hbottom = 4. 623 m