  Question

The volume of a cube is increasing at the rate of $$8 cm^3/s$$. How fast is the surface area increasing when the length of an edge is $$12 cm$$?

Solution

Let $$x$$ be the length of a side, $$V$$ be the volume, and $$S$$ be the surface area of the cube.Then, $$V =x^3$$  and $$S = 6x^2$$  It is given that $$\cfrac {dV}{dt}=8 cm^3/s$$.Then, by using the chain rule, we have: $$\therefore\displaystyle 8=\frac {dV}{dt}=\frac {d}{dt}(x^3)\cdot \frac{d}{dx}=3x^2\cdot \frac {dx}{dt}$$$$\Rightarrow\displaystyle \frac {dx}{dt}=\frac {8}{3x^2}.........(1)$$Now, $$\displaystyle \frac {dS}{dt}=\frac {d}{dt}(6x^2)\cdot \frac{d}{dx}=(12x)\cdot \frac {dx}{dt}$$ [By chain rule]$$\displaystyle =12x\cdot \frac {dx}{dt}=12x\cdot \left ( \frac {8}{3x^2} \right )=\frac {32}{x}$$Thus, when $$x = 12$$ cm,  $$\displaystyle \frac {dS}{dt}=\frac {32}{12} cm^2/s=\frac {8}{3} cm^2/s.$$Hence, if the length of the edge of the cube is $$12$$ cm, then the surface area is increasing  at the rate of $$\cfrac {8}{3} cm^2/s$$.MathematicsRS AgarwalStandard XII

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