Answer :
Answer:
Step-by-step explanation:
We have volume of a cube = [tex]a^3[/tex], where a is the side
Given that [tex]\frac{dV}{dt} =10[/tex]
[tex]V(a) = a^3 \\S(a) =6a^2[/tex], where S is surface area
Differentiate V wrt t
[tex]\frac{dv}{dt} =3a^2 \frac{da}{dt} \\-10 = 3a^2 \frac{da}{dt}[/tex]
To find a: [tex]S= 54 =6a^2\\a= 3[/tex]
Hence we get
[tex]\frac{da}{dt} =\frac{-10}{3a^2} \\=\frac{-10}{27}[/tex]
Hence rate of decrease of surface area = [tex]12a\frac{da}{dt} =12(-\frac{10}{27} )(3) = \frac{-40}{3}[/tex]
Decreasing at the rate of 13.33 m^2/hour.