Answer:
Therefore, the rate of change of the volume of the box when the side length is 10 ft and the height is 12 ft is 680 ft^3/sec.
Explanation:
I'd be happy to explain the solution step by step!
We want to find the rate of change of the volume of a rectangular box with a square base, where the side length is increasing at the rate of 2 ft/sec and the height is decreasing at the rate of 2 ft/sec. We are given that the side length is 10 ft and the height is 12 ft.
Step 1: Write the formula for the volume of the rectangular box
The volume V of a rectangular box with a square base is given by:
V = (side length)^2 x height
Step 2: Take the derivative of the volume formula with respect to time
We want to find the rate of change of the volume with respect to time, so we need to take the derivative of the volume formula with respect to time t. We will use the product rule to differentiate the formula:
dV/dt = 2 x (side length) x (d(side length)/dt) x height + (side length)^2 x (d height/dt)
The first term in the expression represents the rate of change of the volume due to the change in the side length, and the second term represents the rate of change of the volume due to the change in the height.
Step 3: Substitute the given values into the derivative formula
We are given that the side length is increasing at the rate of 2 ft/sec and the height is decreasing at the rate of 2 ft/sec, and that the side length is 10 ft and the height is 12 ft. Substituting these values into the derivative formula, we get:
dV/dt = 2 x 10 x 2 x 12 + 10^2 x (-2)
Simplifying this expression, we get:
dV/dt = 680 ft^3/sec
Therefore, the rate of change of the volume of the box when the side length is 10 ft and the height is 12 ft is 680 ft^3/sec.