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 4 cm, the edge length of the base is increasing at a rate of 2 cm/min, the height of the box is 15 cm, and the height is decreasing at a rate of 3 cm/min.
V = x^2 h
dV/dx = 2 x h
dV/dh = x^2
dV/dt = dV/dx dx/dt + dV/dh dh/dt
x = 4
h = 15
dV/dt = 120 dx/dt + 16 dh/dt
= 120(2) -16(3)
= 240-48
To find the rate at which the volume of the box is changing, we need to differentiate the volume equation with respect to time and then substitute the given values.
Given:
Edge length of the base, x = 4 cm
Rate of change of edge length, dx/dt = 2 cm/min (increasing)
Height of the box, h = 15 cm
Rate of change of height, dh/dt = -3 cm/min (decreasing)
We are looking for the rate of change of volume, dV/dt.
The volume equation is V = x^2h.
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = d/dt (x^2h)
Using the product rule for differentiation:
dV/dt = (d/dt (x^2))h + (d/dt (h))(x^2)
To find each term, we differentiate x^2 and h with respect to time:
d/dt (x^2) = 2x(dx/dt)
d/dt (h) = dh/dt
Now substituting the given values:
x = 4 cm
dx/dt = 2 cm/min
h = 15 cm
dh/dt = -3 cm/min
We can now calculate the rate at which the volume of the box is changing, dV/dt:
dV/dt = (2x(dx/dt))h + (dh/dt)(x^2)
dV/dt = (2 * 4 * 2) * 15 + (-3) * (4^2)
dV/dt = 16 * 15 - 3 * 16
dV/dt = 240 - 48
dV/dt = 192 cm^3/min
Therefore, the rate at which the volume of the box is changing when the edge length of the base is 4 cm and increasing at a rate of 2 cm/min, and the height of the box is 15 cm and decreasing at a rate of 3 cm/min, is 192 cm^3/min.
To find the rate at which the volume of the box is changing, we need to take the derivative of the volume formula with respect to time (t).
Given:
Edge length of the base (x) = 4 cm, and the rate of change of x (dx/dt) = 2 cm/min
Height of the box (h) = 15 cm, and the rate of change of h (dh/dt) = -3 cm/min
The volume of the box is given by V = x^2 * h
Taking the derivative of the volume with respect to time (t), we get:
dV/dt = d/dt (x^2 * h)
Now, let's apply the product rule of differentiation. The derivative of a product of two functions u(t) * v(t) with respect to t is given by:
d/dt (u(t) * v(t)) = u'(t) * v(t) + u(t) * v'(t)
In our case, u(t) = x^2 and v(t) = h. Therefore, u'(t) = d/dt (x^2) and v'(t) = d/dt (h).
Using the product rule, the derivative of the volume with respect to time becomes:
dV/dt = d/dt (x^2 * h) = (d/dt (x^2)) * h + x^2 * (d/dt (h))
Now, let's find the derivatives:
d/dt (x^2) = 2x * dx/dt [Using the power rule of differentiation]
d/dt (h) = dh/dt [Since h is not changing with respect to time]
Substituting these derivatives back into the volume derivative equation:
dV/dt = (2x * dx/dt) * h + x^2 * (dh/dt)
Now, plug in the given values:
x = 4 cm
dx/dt = 2 cm/min
h = 15 cm
dh/dt = -3 cm/min
dV/dt = (2 * 4 * 2) * 15 + (4^2) * (-3)
= 16 * 15 - 16 * 3
= 240 - 48
= 192 cm^3/min
Therefore, the rate at which the volume of the box is changing is 192 cm^3/min.