The surface area of a cube is increasing at the rate of `2 cm^(2)//sec`. When its edge is 90 cm, the volume is increasing at the rate ofA. `1620 cm^(3)//sec`B. `810 cm^(3)//sec`C. `405 cm^(3)//sec`D. `45 cm^(3)//sec`

The surface area of a cube is increasing at the rate of `2 cm^(2)//sec

1.	The surface area of a cube is increasing at the rate of `2 cm^(2)//sec`. When its edge is 90 cm, the volume is increasing at the rate ofA. `1620 cm^(3)//sec`B. `810 cm^(3)//sec`C. `405 cm^(3)//sec`D. `45 cm^(3)//sec`
Answer» Correct Answer - D Let at any time t the length of each edge of the cube be x cm. Let S denote the surface area and V the volume of the cube. Then, `S=6x^(2) and V=x^(3)` `implies (dS)/(dt)=12x (dx)/(dt) and (dV)/(dt)=3x^(2)(dx)/(dt)` `implies 2=12 xx 90(dx)/(dt) and (dV)/(dt)=3xx90^(2)xx(dx)/(dt) [because x=90 cm and (dS)/(dt)=2]` `implies 90xx(dx)/(dt)=(1)/(6) and (dV)/(dt)=3xx90xx(90xx(dx)/(dt))` `implies (dV)/(dt)=3xx90xx(1)/(6) cm^(3)//sec=45 cm^(3)//sec`

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The surface area of a cube is increasing at the rate of `2 cm^(2)//sec`. When its edge is 90 cm, the volume is increasing at the rate ofA. `1620 cm^(3)//sec`B. `810 cm^(3)//sec`C. `405 cm^(3)//sec`D. `45 cm^(3)//sec`

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