Answer :
Answer:
The surface area increases at 12/7 square inches per minute
Step-by-step explanation:
Given
[tex]l = 7[/tex] --- side length
[tex]\frac{dV}{dt} = 3in^3/min[/tex] --- Rate at which volume increases
Required
The rate at [tex]which[/tex] the [tex]surface[/tex] area [tex]increases[/tex]
The volume of a cube is:
[tex]V = l^3[/tex]
Differentiate with respect to time
[tex]\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt}[/tex]
When [tex]l = 7[/tex] and [tex]\frac{dV}{dt} = 3in^3/min[/tex]
The expression becomes:
[tex]3 = 3 * 7^2 \cdot \frac{dl}{dt}[/tex]
Divide both sides by 3
[tex]1 = 7^2 \cdot \frac{dl}{dt}[/tex]
[tex]1 = 49 \cdot \frac{dl}{dt}[/tex]
Divide both sides by 49
[tex]\frac{dl}{dt} = \frac{1}{49}[/tex]
The surface area is calculated as:
[tex]A= 6l^2[/tex]
Differentiate with respect to time
[tex]\frac{dA}{dt} = 12l\frac{dl}{dt}[/tex]
Substitute: [tex]\frac{dl}{dt} = \frac{1}{49}[/tex] and [tex]l = 7[/tex]
[tex]\frac{dA}{dt} = 12 * 7 *\frac{1}{49}[/tex]
[tex]\frac{dA}{dt} = 12 *\frac{1}{7}[/tex]
[tex]\frac{dA}{dt} = \frac{12}{7}[/tex]
The surface area increases at 12/7 square inches per minute