Question

find the dimensions of the rectangular box open at the top of maximum capacity whose surface area is 432 sq cm

Answer 1

Step-by-step explanation:

Let x, y, z respectively be the length, breadth and height of the rectangular box. Since it is open at the top, the surface area (S) is given by S = xy + 2xz + 2yz = 432 (using the data) Volume (V) = xyz We need to find x, y, z such that V is maximum subject to the condition that xy + 2xz + 2yz = 432 Let F = xyz + λ (xy + 2xz + 2yz) We form the equation Fx = 0, F y = 0, Fz = 0 i.e., yz + λ ( y + 2z) = 0 or λ = – yz/( y + 2y) xz + λ (x + 2z) = 0 or λ = – xz/(x + 2z) xy + λ (2x + 2y) = 0 or λ = – xy/2(x + y) Thus the required dimensions are 12, 12, 6.Read more on Sarthaks.com - https://www.sarthaks.com/355157/find-the-dimensions-of-the-rectangular-box-open-at-the-top

Answer 2

Answer:

432 multiple by 4

Step-by-step explanation:

432 ko multiple karo 4 se