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The sum of the surface areas of a cuboid with sides `x ,2x`and `x/3`and a sphere is given to be constant. Prove that the sum of their volumesis minimum, if `x`is equal to threetimes the radius of sphere. Also find the minimum value of the sum of theirvolumes. |
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Answer» Let radius of Sphere in r Surface area=`4pir^2` Surface area of cuboid=`2(lb+bh+hl)` `=2(x*2x+2x*x/3+x*2/3)` `=6x^2` unit `S=4xr^2+6x^2` `r=sqrt((S-6x^2)/(4x)` `S=x*2x*x/3+4/3*x*r^3` `S=2/3X^3+4/3x((S-6x^2)/(6x))^(3/2)` `dv/dx=2x^2+(4x)/(3(4x)^(3/2))*d/dx(5-6x^2)^(3/2)` `2x^2=(12x)/(2(4x)^(1/2))*3/2(5-6x^2)^(1/2)` `x=12/4((S_6x^2)/(4x))` `x=3r,r=x/3` `V=2/3x^3+4/3x(x/3)^3` `V_(min)=2/3x^3+4/8x^3`. |
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