The surface area of a sphere is given by the function

For a radius defined as

The surface area equation becomes

The rate of change of the surface area can then be found by taking the derivative of the equation with respect to time:

The area of a circle is defined by its radius as follows:

In the case of the given function for the radius 

The area is thus

The rate of change can be found by taking the derivative with respect to time:

The area of a right triangle can be written in terms of its legs (the two shorter sides):

For sides  and , the area expression for this problem becomes:

To find where this area has its local maxima/minima, take the derivative with respect to time and set the new equation equal to zero:

At an earlier time , the derivative is postive, and at a later time , the derivative is negative, indicating that  corresponds to a maximum.

To solve for the rate of change of the surface area of the chocolate ball, we must relate that rate to the rate of change of the radius, which we get from the given rate of change of volume:

Now, using the rate of change of volume, solve for the rate of change of the radius:

Now, plug this into the equation for the rate of change of the surface area, using the given radius:

The surface area of a rectangular prism is defined in terms of its sides as follows:

The rate of change of the surface area can then be found by taking the derivative of each side of the equation with respect to time:

For the side functions

And the rate of change of the area becomes:

The area of a rectangle is given by the function:

To compare rates of change between each parameter, take the derivative of each side of this equation with respect to time:

We are told that , , and . If the area is not changing, it follows that .

This reduces the equation to:

A triangle is isosceles when two of its sides are equal. Since the hypotenuse will always be greater than the two legs, this question is essentially considering the moment in time when the two legs are equal.

The lengths of the legs at a moment in time can be defined by the functions:

To find when they sides are equal, set their respective equations equal to each other:

At this time 

The area of a right triangle is given by the formula

So the rate of change of the area can be found by taking the derivative of each side of the equation with respect to time:

The hypotenuse of a right triangle, in terms of the triangle's legs, is given by the Pythagorean theorem as:

For legs  and , the hypotenuse of this problem's triangle can be written as:

To find where this function has maxima/minima, take the derivative with respect to time and set it equal to zero:

The derivative is negative at an earlier time and positive at a later time, indicating a minimum.

The area of a square in terms of its sides is given by the function:

For sides expressed as the function  the area is then

The maximum of the area can be found by taking the derivative of the area function and seeing where it is equal to zero:

To find the derivative use the chain rule and power rule.

Chain Rule: 

Power Rule: 

So there are two times where the rate of change of the area is zero (dismissing negative times):

However, after time , the sides of the square take on negative values, which is geometrically impossible.

 corresponds to the time when the area of the square reaches a maximum value of

Another method of solving this particular problem would be to find the time when the sides achieve their maximum value and then squaring this value.

The area of a rectangle can be expressed in terms of its sides as:

For sides  and , the expression in this problem becomes:

The time(s) where the area reaches a maximum or minimum can then be found by taking the derivative of this expression with respect to time:

To find the derivative use the power rule.

Power Rule: 

There are two times when this expession is equal to zero, which can be found using the quadratic formula:

Note that at time , the length takes a negative value which is impossible. Before time  the derivative is positive, and immediately afterwards the derivative is negative, so it indicates a maximum.