1.

For the box to satisfy certain requirements, its length must be three meter greater than the width, and its height must be two meter less than the width. If it must have a volume of 18 unit, what must be its length ?

Answer»

6 unit
3 unit
4 unit
2 unit

Solution :We have
`V(x)=x^(3)+x^(2)-6`
`18=x^(3)+x^(2) -6X`
`x^(3) +x^(2) -6x -18=0`
`x^(3) -3X^(2) +4x^(2) -12x +6x -18=0`
`x^(2)(x-3) +4x(x-3) +6(x-3) =0`
`(x-3) (x^(2) +4x +6)=0`
Thus WIDTH is 3 unit.
Length `=x+3=6m`
Thus (a) is CORRECT option.


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