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#### karush

##### Well-known member

- Jan 31, 2012

- 2,817

The volume of a cube is increasing at a rate of $$\displaystyle\frac{10 cm^3}{\text {min}}$$

How fast is the surface area increasing when the lenght of an edge is $$\displaystyle 30 \text{ cm}$$

The answer to this is $$\displaystyle \frac{4}{3} \text{ cm}^2$$

$$V=x^3$$ and $$SA=6x^2$$ and $$\displaystyle\frac {dV}{dt} = \displaystyle\frac{10 cm^3}{\text {min}}$$

actually I don't know how to set this up relating $SA$ to $V$

How fast is the surface area increasing when the lenght of an edge is $$\displaystyle 30 \text{ cm}$$

The answer to this is $$\displaystyle \frac{4}{3} \text{ cm}^2$$

$$V=x^3$$ and $$SA=6x^2$$ and $$\displaystyle\frac {dV}{dt} = \displaystyle\frac{10 cm^3}{\text {min}}$$

actually I don't know how to set this up relating $SA$ to $V$

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