Can You Bake the Optimal Cake Using Calculus?

Problem Description

FiveThirtyEight’s Oliver Roeder challenged readers to maximize the volume of a cake comprised of three cylindrical tiers subject to the constraint that the cake must fit beneath a glass cone of a prescribed size in The Riddler:

A mathematician who has a birthday coming up asks his students to make him a cake. He is very particular (he is a constructive set theorist, which explains a lot) and asks his students to make him a three-layer cake that fits under a hollow glass cone he has on his desk. (The cone was given to him as a prize for proving an obscure theorem long ago.) He requires that the cake fill up the maximum amount of volume under the cone as possible and that the layers of the cake have straight vertical sides. (Again, the guy’s particular.) What are the thicknesses of the three layers of the cake in terms of the height of the glass cone? What percentage of the cone’s volume does the cake fill?

Here’s an animation of what the cake looks like. Our goal is to adjust the height/radius of each of the three tiers in order to maximize the volume of the cake:

Drawing a Diagram

While there may be an elegant solution to this problem that uses only elementary techniques, this problem immediately stood out to me as one that could be solved using multivariable calculus optimization. Let’s start with a cone of height $h$ and radius $r$ . Let the height of each tier of the cake be $h_1$ , $h_2$ , and $h_3$ . Also let the remaining space between the third tier of the cake and the top of the cone be $h_0 = h - h_1 - h_2 - h_3$ . Finally, let the radius of each tier of the cake be $r_1$ , $r_2$ , and $r_3$ . Here’s a diagram:

A Formula for the Volume of the Cake

The volume of the entire cone is just $V_{cone} = \frac{\pi r^2 h}{3}$ . The combined volume of the three-tiered cake is a bit more complicated since it is the sum of the volumes of the individual cylindrical tiers, $V_{cake} = \pi r_3^2 h_3 + \pi r_2^2 h_2 + \pi r_1^2 h_1$ . At first, maximizing $V_{cake}$ might seem very difficult since there are so many variables in the equation. Luckily, the diagram above reveals a set of four similar right triangles, which we can use to our advantage:

The first right triangle is the right side of the entire big triangle with a height of $h$ and a radius of $r$ . There’s a second triangle of height $h_0 + h_1 + h_2$ and radius $r_3$ that is formed by the space above the bottom tier of the cake. A third one has a height of $h_0 + h_1$ and a radius of $r_2$ that is in the space above the second tier of the cake. Finally, a fourth one has a height of $h_0$ and a radius of $r_1$ . This leads us to the following proportions:

$\frac{h}{r} = \frac{h_0 + h_1 + h_2}{r_3} = \frac{h_0 + h_1}{r_2} = \frac{h_0}{r_1}$

These proportions can be used to obtain the following equalities:

$r_1 = \frac{h_0 r}{h}$

$r_2 = \frac{(h_0 + h_1) r}{h}$

$r_3= \frac{(h_0 + h_1 + h_2) r}{h}$

If you plug these back into $V_{cake} = \pi r_3^2 h_3 + \pi r_2^2 h_2 + \pi r_1^2 h_1$ , you get the following volume equation as a function of $h_1$ , $h_2$ , and $h_3$ :

$V_{cake}(h_1, h_2, h_3) = \dfrac{\pi r^2}{h^2} \Big (h_1(h - h_1 - h_2 - h_3) + h_2 (h - h_2 - h_3) + h_3(h-h_3) \Big )$

Now remember that $h_1$ , $h_2$ , and $h_3$ are variables while $h$ and $r$ are constants. We can find the gradient of this function and set it equal to the zero vector to find candidates for extrema. I personally found my choice of variables distracting at this point in the problem, so I rewrote my equation as follows using $r = a$ , $h = b$ , $h_1 = x$ , $h_2 = y$ , and $h_3 = z$ :

$V_{cake}(x, y, z) = \dfrac{\pi a^2}{b^2} \Big (x(b - x - y - z) + y (b - y - z) + z(b-z) \Big )$

Maximize the Volume Using the Gradient

This equation is perfect for finding the gradient, $\nabla V_{cake}(x,y,z)$ :

$\nabla V_{cake}(x,y,z) = \dfrac{\pi a^2}{b^2} \Big ( (b-x-y-z)^2 - 2 x(b -x -y - z), (b-y-z)^2-2x(b - x - y - z) - 2y(b - y - z),(b-z)^2-2x(b - x - y - z) - 2y(b - y - z) - 2z(b-z) \Big )$

As usual, set $V_{cake}(x, y, z) = (0,0,0)$ and solve. We get four solutions, the first two of which are extraneous if we require that the cake have three tiers of nonzero volume:

$(x,y,z) = (0, 0, b)$

$(x,y,z) = \Big (0, \dfrac{2b}{3} ,\dfrac{b}{3} \Big )$

$(x,y,z) =\Big (\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} \Big )$

$(x,y,z) = \Big (\dfrac{12b}{23} ,\dfrac{6b}{23} ,\dfrac{5b}{23} \Big )$

Test Candidates Using the Hessian Determinant

It’s a bit of a pain, but you should now spend a few minutes verifying which of these solutions is our desired maximum. This is a perfect time to break out the second derivative test for a function of the three variables. This requires us to compute the Hessian determinant. For a function $f(x,y,z)$ , this means we need to compute the following determinants:

$D_1(x,y,z) = f_{xx}$
$D_2(x,y,z) = \left| \begin{array}{ cc} f_{xx}&f_{xy} \\f_{yx}&f_{yy} \end{array} \right|$
$D_3(x,y,z) = \left| \begin{array}{ ccc} f_{xx}&f_{xy}&f_{xz} \\f_{yx}&f_{yy}&f_{yz} \\ f_{zx}&f_{zy}&f_{zz} \end{array} \right|$

Now we know that $f(x,y,z)$ has a local maximum at $(x_0,y_0,z_0)$ if $D_1( x_0 ,y_0 , z_0 ) \textless 0$ , $D_2( x_0 ,y_0 , z_0 ) \textgreater 0$ , and $D_3( x_0 ,y_0 , z_0 ) \textless 0$ .

Further, $f(x,y,z)$ has a local minimum at $(x_0,y_0,z_0)$ if $D_1( x_0 ,y_0 , z_0 ) \textgreater 0$ , $D_2( x_0 ,y_0 , z_0 ) \textgreater 0$ , and $D_3( x_0 ,y_0 , z_0 ) \textgreater 0$ .

Otherwise, $f(x,y,z)$ has a saddle point at $(x_0,y_0,z_0)$ .

Testing the candidate $(x,y,z) =\Big (\dfrac{12b}{23} ,\dfrac{6b}{23} ,\dfrac{5b}{23} \Big )$ gives us a saddle point:
$D_1 \Big (\dfrac{12b}{23} ,\dfrac{6b}{23} ,\dfrac{5b}{23} \Big ) = \dfrac{24 a^2 \pi}{23b} \textgreater 0$
$D_2 \Big (\dfrac{12b}{23} ,\dfrac{6b}{23} ,\dfrac{5b}{23} \Big ) = - \dfrac{864 a^2 \pi^2}{529 b^2} \textless 0$
$D_3 \Big (\dfrac{12b}{23} ,\dfrac{6b}{23} ,\dfrac{5b}{23} \Big ) = \dfrac{1728 a^6 \pi^3}{529 b^3} \textgreater 0$

On the other hand, $\Big(\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} \Big )$ gives us our desired local maximum:
$D_1 \Big (\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} \Big ) =-\dfrac{552 \pi a^2}{421 b} \textless 0$
$D_2 \Big (\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} \Big ) = \dfrac{389344 \pi ^2 a^4}{177241 b^2} \textgreater 0$
$D_3 \Big (\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} \Big ) = -\dfrac{778688 \pi ^3 a^6}{177241 b^3} \textless 0$

The Solution

Now that we have our local maximum at $\Big (\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} \Big )$ , let’s plug this ordered triple back into $V_{cake}(x, y, z) = \dfrac{\pi a^2}{b^2} \Big (x(b - x - y - z) + y (b - y - z) + z(b-z) \Big )$ . Since we are done using calculus at this point, let’s also switch back to $a = r$ , $b = h$ at the same time:

$V_{cake} \Big (\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} \Big ) = \dfrac{1119364 \pi r^2 h}{4785507}$

This is the maximum volume of our three-tiered cake. And here’s the maximum percentage of the cone filled with cake:

$\dfrac{V_{cake}(\dfrac{92b}{421} ,\dfrac{230b}{1263} ,\dfrac{205b}{1263} )}{V_{cone}} \approx 70.1721 \%$

Mathematica Demo

I created a Mathematica demo of my solution, which you can preview below as an animated GIF. If you would like, you can also download my full Mathematica notebook (either LayerCakeProblem.nb or LayerCakeProblem.cdf).

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