an-open-rectanqular-container-is-to-have-a-volume-of-62-5cm-3-find-the-least-possible-surface-area-of-the-material-required-

Question Number 115368 by mathdave last updated on 25/Sep/20

a n o p e n r e c t a n q u l a r c o n t a i n e r i s t o

h a v e a v o l u m e o f 62.5 {c m}^{3} . f i n d t h e l e a s t

p o s s i b l e s u r f a c e a r e a o f t h e m a t e r i a l

r e q u i r e d

Answered by 1549442205PVT last updated on 25/Sep/20

Denote the sizes of bottom of the

o p e n r e c t a n q u l a r c o n t a i n e r by x, y;

let z is the heigh of the container .

From the hypothesis we have

xyz = 62.5 ({cm}^{3}) . Then the surface area

the o p e n r e c t a n q u l a r c o n t a i n e r is

S = 2 xz + 2 yz + xy . Applying {Cauchy}^{'} s

inequality for three positive numbers

we get

S = 2 xz + 2 yz + xy ⩾ 3^{3} \sqrt{2 xz . 2 yz . xy}

= 3^{3} \sqrt{4 {(xyz)}^{2}} = 3^{3} \sqrt{4.62 . 5^{2}} = 75

The equality ocurrs if and only if

{\begin{cases} xyz = 62.5 \\ 2 xz = 2 yz = xy \end{cases} \Leftrightarrow x = y = 5, z = \frac{5}{2}

Thus, the least surface area of the

container equal to {75 cm}^{2} when

x = y = 5 cm, z = 2 . 5 cm

Commented by mathdave last updated on 25/Sep/20

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