How to Maximize the Volume of a Cone with Fixed Slant Height

A cone’s volume cannot be maximized without specifying what is held fixed: increasing both its radius and height increases its volume without limit. A common version of this problem fixes the slant height of a right circular cone. This article solves that version and explains why a different constraint can lead to a different answer.

Model the cone

Let $r$ be the base radius, $h$ the perpendicular height, and $l$ the slant height. The cone’s volume is

$$V=\frac{1}{3}\pi r^2h.$$

A cross-section through the axis forms a right triangle, so the dimensions satisfy

$$r^2+h^2=l^2.$$

With $l$ fixed, the radius and height cannot vary independently. Use the constraint to eliminate $r$:

$$r^2=l^2-h^2.$$

Thus the volume, as a function of height alone, is

$$V(h)=\frac{\pi}{3}(l^2-h^2)h =\frac{\pi}{3}(l^2h-h^3), \qquad 0\le h\le l.$$

Find and check the critical point

Differentiate with respect to $h$, treating $l$ as a constant:

$$V'(h)=\frac{\pi}{3}(l^2-3h^2).$$

Set the derivative equal to zero:

$$l^2-3h^2=0 \quad\Longrightarrow\quad h=\frac{l}{\sqrt{3}}.$$

The negative root is not a possible cone height. Since

$$V''(h)=-2\pi h,$$

we have $V''(l/\sqrt{3})<0$, confirming a local maximum. It is also the global maximum on the allowed interval: the volume is zero at both endpoints ($h=0$ and $h=l$), while it is positive at the critical point.

Use the constraint to find the radius at this maximum:

$$r^2=l^2-\frac{l^2}{3}=\frac{2l^2}{3}, \qquad r=l\sqrt{\frac{2}{3}}.$$

Therefore the dimensions satisfy $h/r=1/\sqrt{2}$. Substituting them into the volume formula gives

$$V_{\max} =\frac{1}{3}\pi\left(\frac{2l^2}{3}\right)\left(\frac{l}{\sqrt{3}}\right) =\frac{2\pi l^3}{9\sqrt{3}}.$$

Worked example

Suppose the slant height is $l=6$ units. The volume-maximizing height and radius are

$$h=\frac{6}{\sqrt{3}}=2\sqrt{3}, \qquad r=6\sqrt{\frac{2}{3}}=2\sqrt{6}.$$

The maximum volume is

$$V_{\max}=\frac{2\pi(6)^3}{9\sqrt{3}} =16\sqrt{3}\pi \approx 87.1\text{ cubic units}.$$

As a check, the dimensions satisfy $r^2+h^2=24+12=36=l^2$, and the computed volume agrees with $V=\frac{1}{3}\pi r^2h$.

If the fixed quantity is different

Do not use the result above automatically if the problem fixes another measurement. For example, if the lateral surface area $A=\pi r l$ is fixed instead, maximizing the volume gives a different proportion: $h/r=\sqrt{2}$, not $1/\sqrt{2}$. The constraint changes the one-variable volume function, so it changes the optimum.

Read the problem statement for the fixed quantity—such as slant height, surface area, or a sphere containing the cone—before differentiating. If no constraint is given, there is no finite maximum to calculate.

Common mistakes

• Optimizing without a constraint. The formula $V=\frac{1}{3}\pi r^2h$ alone does not determine a largest cone; the dimensions need a restriction.
• Confusing height with slant height. The constraint $r^2+h^2=l^2$ uses the perpendicular height and the slant height, not the radius and the slant height alone.
• Maximizing the wrong variable expression. Substitute $r^2=l^2-h^2$ into the volume formula; do not treat $r$ and $h$ as independent after fixing $l$.
• Ignoring the domain. A physical cone has $0\le h\le l$ and nonnegative radius. The stationary point must also be checked against the endpoints.
• Reporting a ratio without naming the constraint. The optimal height-to-radius ratio depends on what is fixed, so include that condition in the answer.