Respuesta :

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.