A cube has a total volume of 120 m^3. Its length must be three times as long as its width. Find the dimensions of such a cube with the lowest surface area.
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Expert's answer
2012-12-11T09:34:26-0500
Surface area S=2*(a*b+b*c+a*c)=(from the taska=3b)=2*(9b^2+4b*c) V=a*b*c=120, so c=120/(a*b)=40/b^2 So, S=2*(9b^2+160/b) It's derivative is:s'=36b-160/b^2 Minimum is at point s'=0 36b=320/b^2 B^3=320/36=80/9=> b=2SQRT3(10/9) a=3b=6sqrt3(80/9) c=120/(a*b)=120/12sqrt3(10/9)^2=10*sqrt3(81/100)=30*sqrt3(0.03)
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