At a certain instant the edge of a cube is 7 inches long and the volume is increasing at the rate of 3 cubic inches per minute. How fast is the surface area of the cube increasing

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

Use the distributive property to remove the parentheses.
(w - 12) 11

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