## liizzyliizz 4 years ago The volume of a cube is increasing at a a rate of 300 in^3/min. at the instant when the edge is 20 inches at what rate is the edge changing.

1. anonymous

edge is $x = \sqrt[3]{V}$ so $x'(t)=\frac{1}{3\sqrt[3]{V^2}}\times V'(t)$

2. anonymous

you are told that $V'(t)=300$ so you get $x'(t)=\frac{300}{3\sqrt[3]{V^2}}=\frac{100}{\sqrt[3]{V^2}}$ so all you need to finish is to find out what V is when x is 20

3. liizzyliizz

4. liizzyliizz

I still blame my algebra skills, but that is just me lol

5. lgbasallote

dV/dT = 300 when e = 20 volume of a cube = e^3 dV/dT = 3e^2(de/dt) 300 = 3e^2 (de/dt) de/dt = 300/3e^2 when e =20 de/dt = 300/3(20)^2 de/dt = 1/4 in/min is the answer I came up with.

6. liizzyliizz

^ that made a lot more sense lol