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.
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To find the rate at which the volume of the box is changing, we need to differentiate the volume function with respect to time (t) and then plug in the given values.
Given:
Edge length of the base (x) = 10 cm
Rate of change of the edge length of the base (dx/dt) = 3 cm/min (positive rate indicates an increase)
Height (h) = 5 cm
Rate of change of the height (dh/dt) = -1 cm/min (negative rate indicates a decrease)
Volume of the box (V) = x^2 * h
Differentiating both sides with respect to time (t), we get:
dV/dt = 2x * dx/dt * h + x^2 * dh/dt
Plugging in the given values, we have:
x = 10 cm
dx/dt = 3 cm/min
h = 5 cm
dh/dt = -1 cm/min
Substituting these values into the equation, we get:
dV/dt = 2(10 cm) * (3 cm/min) * (5 cm) + (10 cm)^2 * (-1 cm/min)
Simplifying, we get:
dV/dt = 60 cm^2/min + 100 cm^2/min
Combining like terms, we get:
dV/dt = 160 cm^2/min
Therefore, the rate at which the volume of the box is changing when the edge length of the base is 10 cm and increasing at a rate of 3 cm/min, and the height of the box is 5 cm and decreasing at a rate of 1 cm/min is 160 cm^2/min.
To find the rate at which the volume of the box is changing, we need to use the chain rule and differentiate the volume formula with respect to time.
Given:
- Edge length of the base, x = 10 cm
- Rate at which the edge length of the base is increasing, dx/dt = 3 cm/min
- Height of the box, h = 5 cm
- Rate at which the height is decreasing, dh/dt = -1 cm/min
The volume of the box is given by 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 of differentiation, we have:
dV/dt = (2x)(dx/dt)h + x^2(dh/dt)
Substituting the given values for x, dx/dt, and dh/dt, we find:
dV/dt = (2 * 10)(3)(5) + (10^2)(-1)
Simplifying, we get:
dV/dt = 300 - 100
dV/dt = 200 cm^3/min
Therefore, the rate at which the volume of the box is changing is 200 cm^3/min.
v = x^2 h
now use the product rule and chain rule:
dv/dt = 2xh dx/dt + x^2 dh/dt
Now just plug in your numbers