Question

A open box with a square base is to be made. Its total surface area is c², a constant. Prove that its maximum volume is c³/6√3.

Answer 1

Answer:

c³/6√3

Step-by-step explanation:

Let say open box base size is  b * b

& height = h

Then Surface area =  b² + 4bh  = c²

=> h = (c² - b²)/4b

Volume = b²h  = b² (c² - b²)/4b

= b (c² - b²)/4

= bc²/4 - b³/4

dV/db = c²/4 - 3b²/4

if dV/db = 0 then

c²/4 - 3b²/4 = 0

=> c² = 3b²

=> b = c/√3

d²V/db² = - 6b/4 is - ve so volume is maximum at b = c/√3

b = c/√3

h =  (c² - b²)/4b = (c² - c²/3)/(4c/√3)  = (2c²/3)/(4c/√3))

= c√3/6

Volume = b²h  = (c²/3)( c√3/6)  = c³/6√3

Hence proved that max voulme =  c³/6√3