# Answer to Question #20347 in Calculus for Sara

Question #20347

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.

Expert's answer

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)

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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