Question

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.

Answer 1

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

Answer 2

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

Answer 3

Ahh, so for future problems similar to this how would one know to go about this.

Answer 4

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

Answer 5

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.

Answer 6

^ that made a lot more sense lol