an open rectanqular container is to have a volume of 62.5cm^3 .find the least possible surface area of the material required


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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
	$Denote the sizes of bottom of the open rectanqular container by x,y; let z is the heigh of the container. From the hypothesis we have xyz=62.5(cm^3 ).Then the surface area the open rectanqular container is S=2xz+2yz+xy.Applying Cauchy′s inequality for three positive numbers we get S=2xz+2yz+xy≥3^3 (√(2xz.2yz.xy)) =3^3 (√(4(xyz)^2 ))=3^3 (√(4.62.5^2 ))=75 The equality ocurrs if and only if { ((xyz=62.5)),((2xz=2yz=xy)) :}⇔x=y=5,z=(5/2) Thus,the least surface area of the container equal to 75cm^(2 ) when x=y=5cm,z=2.5cm$
	$Denote the sizes of bottom of the$ $o p e n r e c t a n q u l a r c o n t a i n e r by x, y;$ $let z is the heigh of the container .$ $From the hypothesis we have$ $xyz = 62.5 ({cm}^{3}) . Then the surface area$ $the o p e n r e c t a n q u l a r c o n t a i n e r is$ $S = 2 xz + 2 yz + xy . Applying {Cauchy}^{'} s$ $inequality for three positive numbers$ $we get$ $S = 2 xz + 2 yz + xy ⩾ 3^{3} \sqrt{2 xz . 2 yz . xy}$ $= 3^{3} \sqrt{4 {(xyz)}^{2}} = 3^{3} \sqrt{4.62 . 5^{2}} = 75$ $The equality ocurrs if and only if$ ${\begin{cases} xyz = 62.5 \\ 2 xz = 2 yz = xy \end{cases} \Leftrightarrow x = y = 5, z = \frac{5}{2}$ $Thus, the least surface area of the$ $container equal to {75 cm}^{2} when$ $x = y = 5 cm, z = 2 . 5 cm$
	Commented by mathdave last updated on 25/Sep/20

	$g u d w o r k$
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