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

Answer:

$\frac{dA}{dt}=\frac{6}{5}cm^2/min$

Step-by-step explanation:

Given

$\frac{dV}{dt}=12cm^3/min$ <-- The change in volume with respect to time

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

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

$\frac{dA}{dt}=?$ <-- The change in surface area with respect to time

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

Solve for ds/dt:

$V=s^3$

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

$12=3(40)^2\frac{ds}{dt}$

$12=4800\frac{ds}{dt}$

$\frac{1}{400}=\frac{ds}{dt}$

Solve for dA/dt:

$A=6s^2$

$\frac{dA}{dt}=12s\frac{ds}{dt}$

$\frac{dA}{dt}=12(40)(\frac{1}{400})$

$\frac{dA}{dt}=12(\frac{1}{10})$

$\frac{dA}{dt}=\frac{12}{10}$

$\frac{dA}{dt}=\frac{6}{5}cm^2/min$

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.

Let me know if you need more help with related rates!