Video Transcript
Given that the sum of the surface
areas of a sphere and a right circular cylinder is 1000𝜋 centimeters squared, and
their radii are equal, find the radius of the sphere that makes the sum of their
volume at its maximum value.
So, in this question, we’re asked
to maximize the sum of the volumes of two three-dimensional solids, subject to a
constraint about the sum of their surface areas. Let’s begin by writing down formulae
for the surface areas of both the sphere and the right circular cylinder. And as their radii are the same, we
can use the same letter 𝑟 for both. For the sphere, first of all, its
surface area is given by four 𝜋𝑟 squared. For the cylinder, its surface area
is two 𝜋𝑟 squared plus two 𝜋𝑟ℎ, where ℎ represents the height of the
cylinder.
As the sum of these surface areas
is 1000𝜋 centimeters squared, we can form an equation, four 𝜋𝑟 squared plus two
𝜋𝑟 squared plus two 𝜋𝑟ℎ equals 1000𝜋. We can then combine the like terms
on the left-hand side and then divide through by a 𝜋, as this is a common factor in
all terms. We could also divide through by
two, as all the coefficients are even, to give three 𝑟 squared plus 𝑟ℎ equals
500. We can’t do anything further with
this equation at this point, as we have two unknowns 𝑟 and ℎ. So, next, we recall the formulae
for the volume of a sphere and the volume of a cylinder.
The volume of a sphere is
four-thirds multiplied by 𝜋 multiplied by its radius cubed. And the volume of a cylinder is 𝜋
multiplied by its radius squared multiplied by its height. So, we have that the total volume
of these two solids is four-thirds 𝜋𝑟 cubed plus 𝜋𝑟 squared ℎ. We want to maximize the sum of
these volumes, 𝑉 total. Now, this will be maximized when
its rate of change with respect to either 𝑟 or ℎ is equal to zero, which will be
when its first derivative is equal to zero. But before we can differentiate, we
need to express 𝑉 total in terms of a single variable.
It’s much more straightforward to
rearrange our surface area constraint to give an expression for ℎ in terms of 𝑟
than it is to give an expression for 𝑟 in terms of ℎ. We have ℎ equals 500 minus three 𝑟
squared over 𝑟. We can then substitute this
expression for ℎ into our expression for the total volume so that it is in terms of
𝑟 only. We can cancel a factor of 𝑟 in the
second term and then distribute the parentheses to give four-thirds 𝜋𝑟 cubed plus
500𝜋𝑟 minus three 𝜋𝑟 cubed. We have an expression for 𝑉 total
in terms of 𝑟 only.
Next, we need to find the first
derivative d𝑉 total by d𝑟, so we’ll create a little bit of space in order to do
this. By applying the power rule of
differentiation, we see that the derivative of 𝑉 total with respect to 𝑟 is equal
to four-thirds 𝜋 multiplied by three 𝑟 squared plus 500𝜋 minus three 𝜋
multiplied by three 𝑟 squared, which all simplifies to 500𝜋 minus five 𝜋𝑟
squared. Next, in order to find critical
points, we need to set this derivative equal to zero and solve for 𝑟.
We can divide through by five 𝜋,
giving zero equals 100 minus 𝑟 squared. Adding 𝑟 squared to both sides
gives 𝑟 squared equals 100. And we then find 𝑟 by square
rooting. We only need to take the positive
square root as the radius of a solid must be a positive value. So, we see that 𝑟 is equal to
10. We now know that the combined
volume of these two solids has a critical point when the radius is equal to 10. But we must now confirm that it is
a maximum.
We perform the second derivative
test. Differentiating our expression for
d𝑉 total by d𝑟 again, with respect to 𝑟, gives negative 10𝜋𝑟. And evaluating this when 𝑟 is
equal to 10 gives negative 100𝜋. This is negative, which confirms
that the critical point is indeed a maximum. So, we found that the radius of the
sphere and also the radius of the right circular cylinder which maximizes the sum of
their volumes, subject to the given surface area constraint, is 10 centimeters.
