A box with square base and no top is to hold a volume of V. Find in terms of V the dimensions of the box that requires the least material for the 5 sides.Also find the ratio of height to the side of the base (this ratio will not involve V).

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Solution

$L e t t h e h e i g h t o f t h e b o x b e h a n d s i d e o f s q u a r e b a s e b e a V = A r e a o f b a s e \times h e i g h t V = a^{2} h h = \frac{V}{a^{2}} I f s u r f a c e a r e a o f t h e b o x i s l e a s t t h e n t h e a m o u n t o f m a t e r i a l r e q u i r e d t o m a k e i t w i l l b e t h e l e a s t S u r f a c e a r e a o f b o x, S = L a t e r a l s u r f a c e a r e a o f a c u b o i d + a r e a o f b a s e \{l = b = a\} S = 2 (a + a) h + a^{2} S = 4 a h + a^{2} S = 4 a \frac{V}{a^{2}} + a^{2} S = \frac{4 V}{a} + a^{2} F o r m a x i m a o r m i n i m a \frac{d S}{d a} = 0 \frac{d S}{d a} = - \frac{4 V}{a^{2}} + 2 a = 0 \frac{2 V}{a^{2}} = a a^{3} = 2 V a = {(2 V)}^{\frac{1}{3}} \frac{d^{2} V}{d a^{2}} = \frac{8 V}{a^{3}} + a > 0 T h e r e f o r e, V h a s m i n i m a h = \frac{V}{a^{2}} = \frac{V}{{(2 V)}^{\frac{2}{3}}} = \frac{V^{\frac{1}{3}}}{2^{\frac{2}{3}}}$

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