The volume of a cube is increasing at the rate of 1200 cm supmin at the instant its edges are 20 cm long. At what rate are the edges changing at that instant?

The volume of a cube with edge length ee is given by: V=e3 Differentiate with respect to t: dVdt=3e2(dedt) Substitute dVdt=1200 cm3min and e=20 cm then solve for dedt: 1200=3(20)2⋅(dedt) 1200=1200(dedt) 1=dedt or dedt=1cmmin

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Efan Halliday

Answered question

2021-05-19

The volume of a cube is increasing at the rate of $1200 c m \frac{s u p}{m i n}$ at the instant its edges are 20 cm long. At what rate are the edges changing at that instant?

Answer & Explanation

aprovard

Skilled2021-05-20Added 94 answers

The volume of a cube with edge length ee is given by:
$V = e^{3}$
Differentiate with respect to t: $\frac{d V}{d t} = 3 e^{2} (\frac{d e}{d t})$
Substitute $\frac{d V}{d t} = 1200 \frac{c m^{3}}{m i n}$ and $e = 20$ cm then solve for $\frac{d e}{d t} :$ $1200 = 3 {(20)}^{2} \cdot (\frac{d e}{d t})$
$1200 = 1200 (\frac{d e}{d t})$
$1 = \frac{d e}{d t}$
or
$\frac{d e}{d t} = 1 \frac{c m}{min}$

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