A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x2h cm3. Find the rate at which the volume of the box is changing when the edge length of the base is 10 cm, the edge length of the base is increasing at a rate of 3 cm/min, the height of the box is 5 cm, and the height is decreasing at a rate of 1 cm/min.

i have no idea

you must have meant:

V = x^2 h
dV/dt = x^2 dh/dt + 2xh dx/dt

now just fill in the given data

2000

To find the rate at which the volume of the box is changing, we need to take the derivative of the volume function with respect to time and evaluate it at the given values of the variables.

Given:
Edge length of the base, x = 10 cm (constant)
Rate of change of the edge length of the base, dx/dt = 3 cm/min (given as positive because it is increasing)
Height of the box, h = 5 cm (constant)
Rate of change of the height, dh/dt = -1 cm/min (given as negative because it is decreasing)

The volume of the box is given by V = x^2h cm^3.

Taking the derivative of V with respect to time, we apply the product rule:

dV/dt = (2xh) (dx/dt) + (x^2) (dh/dt)

Substituting the given values:

dV/dt = (2 * 10 * 5) (3) + (10^2) (-1)

Simplifying:

dV/dt = 100 (3) - 100 (-1)

dV/dt = 300 + 100

dV/dt = 400 cm^3/min

Therefore, the volume of the box is changing at a rate of 400 cm^3/min.