Question

The volume of a cube is increasing at a rate of 12 cm3/min. How fast is the surface area increasing when the length of an edge is 40 cm

Answer 1

`(dA)/(dt)=(6)/(5)cm^2/min`

Explanation:

Given

`(dV)/(dt)=12cm^3/min` <-- The change in volume with respect to time

`V=s^3` <-- Volume of a cube

`A=6s^2` <-- Surface area of a cube

`(dA)/(dt)=?` <-- The change in surface area with respect to time

`s=40cm` <-- Length of edge of cube

Solve for ds/dt:

`V=s^3`

`(dV)/(dt)=3s^2(ds)/(dt)`

`12=3(40)^2(ds)/(dt)`

`12=4800(ds)/(dt)`

`(1)/(400)=(ds)/(dt)`

Solve for dA/dt:

`A=6s^2`

`(dA)/(dt)=12s(ds)/(dt)`

`(dA)/(dt)=12(40)((1)/(400))`

`(dA)/(dt)=12((1)/(10))`

`(dA)/(dt)=(12)/(10)`

`(dA)/(dt)=(6)/(5)cm^2/min`

Therefore, the surface area of the cube is increasing at a rate of 6/5 cm²/min when the length of an edge is 40 cm.

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