Answer:
The surface area of the cube decreasing at a rate of 3.33 square inches/min.
Step-by-step explanation:
It is given that, the volume of a cube is decreasing at a rate of 0.5 cubic inches/min.
[tex]\dfrac{dV}{dt}=-0.5\text{ in}^3/min[/tex] ...(1)
We know that, volume of a cube is
[tex]V=a^3[/tex]
Differentiate w.r.t. t.
[tex]\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}[/tex]
[tex]-0.5=3a^2\dfrac{da}{dt}[/tex]
[tex]-\dfrac{0.5}{3a^2}=\dfrac{da}{dt}[/tex] ...(2)
Surface area of the cube is
[tex]A=6a^2[/tex]
Differentiate w.r.t. t.
[tex]\dfrac{dA}{dt}=12a\dfrac{da}{dt}[/tex]
[tex]\dfrac{dA}{dt}=12a\left(-\dfrac{0.5}{3a^2}\right)[/tex] [tex][Using (2)][/tex]
[tex]\dfrac{dA}{dt}=-\dfrac{2}{a}[/tex]
We need to find how fast is the surface area of the cube decreasing when the length of an edge is 0.6 inches.
Substitute a=0.6 in the above equation.
[tex]\dfrac{dA}{dt}=-\dfrac{2}{0.6}[/tex]
[tex]\dfrac{dA}{dt}=-3.33[/tex]
Therefore, the surface area of the cube decreasing at a rate of 3.33 square inches/min.