Question Video: Finding the Radius of a Sphere and a Cylinder That Maximizes the Sum of Their Volumes given the Sum of Their Surface Areas Using Differentiation

Given that the sum of the surface areas of a sphere and a right circular cylinder is 1000πœ‹ cmΒ², and their radii are equal, find the radius of the sphere that makes the sum of their volume at its maximum value.

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Video Transcript

Given that the sum of the surface areas of a sphere and a right circular cylinder is 1000πœ‹ 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 fomulae 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 π‘Ÿ for both. For the sphere, first of all, its surface area is given by four πœ‹π‘Ÿ squared. For the cylinder, its surface area is two πœ‹π‘Ÿ squared plus two πœ‹π‘Ÿβ„Ž, where β„Ž represents the height of the cylinder.

As the sum of these surface areas is 1000πœ‹ centimeters squared, we can form an equation, four πœ‹π‘Ÿ squared plus two πœ‹π‘Ÿ squared plus two πœ‹π‘Ÿβ„Ž equals 1000πœ‹. We can then combine the like terms on the left-hand side and then divide through by a πœ‹, 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 π‘Ÿ squared plus π‘Ÿβ„Ž equals 500. We can’t do anything further with this equation at this point, as we have two unknowns π‘Ÿ and β„Ž. 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œ‹ multiplied by its radius cubed. And the volume of a cylinder is πœ‹ multiplied by its radius squared multiplied by its height. So, we have that the total volume of these two solids is four-thirds πœ‹π‘Ÿ cubed plus πœ‹π‘Ÿ squared β„Ž. 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 π‘Ÿ or β„Ž 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 β„Ž in terms of π‘Ÿ than it is to give an expression for π‘Ÿ in terms of β„Ž. We have β„Ž equals 500 minus three π‘Ÿ squared over π‘Ÿ. We can then substitute this expression for β„Ž into our expression for the total volume so that it is in terms of π‘Ÿ only. We can cancel a factor of π‘Ÿ in the second term and then distribute the parentheses to give four-thirds πœ‹π‘Ÿ cubed plus 500πœ‹π‘Ÿ minus three πœ‹π‘Ÿ cubed. We have an expression for 𝑉 total in terms of π‘Ÿ only.

Next, we need to find the first derivative d𝑉 total by dπ‘Ÿ, 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 π‘Ÿ is equal to four-thirds πœ‹ multiplied by three π‘Ÿ squared plus 500πœ‹ minus three πœ‹ multiplied by three π‘Ÿ squared, which all simplifies to 500πœ‹ minus five πœ‹π‘Ÿ squared. Next, in order to find critical points, we need to set this derivative equal to zero and solve for π‘Ÿ.

We can divide through by five πœ‹, giving zero equals 100 minus π‘Ÿ squared. Adding π‘Ÿ squared to both sides gives π‘Ÿ squared equals 100. And we then find π‘Ÿ 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 π‘Ÿ 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 dπ‘Ÿ again, with respect to π‘Ÿ, gives negative 10πœ‹π‘Ÿ. And evaluating this when π‘Ÿ is equal to 10 gives negative 100πœ‹. 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.

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