Calculus Problem Optimization

A large ice cube is melting, each edge shrinking by 2 centimeters per minute. Find the rate at which the volume is changing at the moment when the length of each edge is 15 cm. Assume the cube maintains its shape throughout the melting process.

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Problem:

A large ice cube is melting, each edge shrinking by 2 centimeters per minute. Find the rate at which the volume is changing at the moment when the length of each edge is 15 cm. Assume the cube maintains its shape throughout the melting process.

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Solution:

Let the length of each edge of the ice cube be denoted by . The volume of the cube, , is given by

V = l^3

Since the edges are shrinking at a rate of 2 centimeters per minute, we write

\frac{dl}{dt} = -2

where represents time. The negative sign is there because is decreasing with time. To find the rate at which the volume change, , we can use the chain rule.

\frac{dV}{dt} = 3l^2 \cdot \frac{dl}{dt} = 3l^2 (-2) = -6l^2

At the moment when the length of each edge is 15 cm, we have . Therefore, the rate at which the volume is changing is

\frac{dV}{dt} = -6(15)^2 = -1350 cm^3/min

In other words, the volume of the ice cube is decreasing at a rate of 1350 cubic centimeters per minute.

Interpretation:

The negative sign in the expression for tells us that the volume is decreasing. The rate of change, cubic centimeters per minute, is quite large. This means that the ice cube is melting very quickly.

Remarks:

In the real world, the melting process would not be as simple as we have described it here. The edges of the ice cube would not shrink at a perfectly uniform rate, and the shape of the cube would change as it melted. However, the mathematical analysis presented here gives us a good approximation of the rate at which the volume of the ice cube is changing.

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