Of-all-rectangular-boxes-without-a-lid-and-having-a-given-surface-area-Find-the-one-with-maximum-volume-

Question Number 6135 by sanusihammed last updated on 15/Jun/16

O f a l l r e c t a n g u l a r b o x e s w i t h o u t a l i d a n d h a v i n g a g i v e n

s u r f a c e a r e a . F i n d t h e o n e w i t h m a x i m u m v o l u m e .

Commented by FilupSmith last updated on 15/Jun/16

Edge lengths a, b, c

Max volume when a = b = c

V = a b c = a^{3} = b^{3} = c^{3}

Answered by Rasheed Soomro last updated on 16/Jun/16

L e t a, b a n d c a r e d i m e n n t i o n s o f

r e c t a n g u l a r b o x .

T h e s u r f a c e a r e a = 2 (a b + b c + c a)

L e t t h e l i d h a s a a n d c d i m e n t i o n s

T h e a r e a o f l i d s u r f a c e = a c

A = W i t h o u t l i d s u r f a c e a r e a = 2 (a b + b c + c a) - a c

= 2 a b + 2 b c + c a

L e t t h e r e i s a c u b i c b o x (w i t h o u t a l i d) o f d i m e n t i o n d

h a v i n g t h e s u r f a c e a r e a e q u a l t o t h a t o f a b o v e b o x .

A = W i t h o u t l i d s u r f a c e a r e a = d^{2} + 2 d^{2} + d^{2} = 5 d^{2}

∴ 2 a b + 2 b c + c a = 5 d^{2}

d = \sqrt{\frac{2 a b + 2 b c + c a}{5}}

N o w {l e t}^{'} s c o m p a r e t h e v o l u m e s o f a b o v e b o x e s

V o l u m e o f r e c t a n g u l a r b o x = a b c

V o l u m e o f c u b i c b o x = d^{3} = {(\frac{2 a b + 2 b c + c a}{5})}^{\frac{3}{2}}

{(\frac{2 a b + 2 b c + c a}{5})}^{\frac{3}{2}} ? a b c \forall a, b, c > 0

A s s u m p t i o n :

{(\frac{2 a b + 2 b c + c a}{5})}^{\frac{3}{2}} ⩾ a b c \forall a, b, c > 0

{C a n}^{'} t c o n t i n u e