Answer:
Therefore the volume of cube is increasing at a rate [tex]\frac{15}{2}[/tex] cubic meter per hour.
Step-by-step explanation:
Given that the surface area of a cube is increasing at a rate 15 square meter per hour.
i.e [tex]\frac{dA}{dt}=15[/tex] square meter per hour.
A = the surface area of the cube
At a certain instant the surface area is 24 square meter.
At that time the side of the cube be x.(let)
The surface area of the cube is =6 x²
According to problem,
6 x² = 24
[tex]\Rightarrow x^2= \frac{24}{6}[/tex]
[tex]\Rightarrow x^2= 4[/tex]
⇒x= 2
At that time the side of the cube was 2 meter.
Again,
A= 6x²
Differentiating with respect to t
[tex]\frac{dA}{dt}= 6.2.x \frac{dx}{dt}[/tex]
[tex]\Rightarrow \frac{dA}{dt}= 12x \frac{dx}{dt}[/tex]
Putting [tex]\frac{dA}{dt}=15[/tex]
[tex]\Rightarrow 15 =12x \frac{dx}{dt}[/tex]
[tex]\Rightarrow \frac{dx}{dt}=\frac{15}{12x}[/tex]
[tex]\therefore \frac{dx}{dt}|_{x=2}=\frac{15}{12\times2}=\frac{5}{8}[/tex]
The volume of the cube is v = x³
v = x³
Differentiating with respect to t
[tex]\frac{dv}{dt}=3x^2\frac{dx}{dt}[/tex]
Putting [tex]\frac{dx}{dt}=\frac{5}{8}[/tex]
[tex]\frac{dv}{dt}=3x^2\times \frac{5}{8}[/tex]
[tex]\Rightarrow \frac{dv}{dt}|_{x=2}=3.2^2\times \frac{5}{8}[/tex]
[tex]=\frac{15}{2}[/tex]
Therefore the volume of cube is increasing at a rate [tex]\frac{15}{2}[/tex] cubic meter per hour.
