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Difficulty: HardStationary Points, Maxima, and Minima

A closed cylindrical metal container has a total surface area of 54π cm254\pi\text{ cm}^2. What radius, in centimeters, of the circular base will yield the maximum volume for the container?

Answer: 3 cm

Answer

The radius of the circular base that maximizes the volume is 3 cm.
To find the radius that yields maximum volume, we first express height hh in terms of radius rr using the total surface area formula 2πr2+2πrh=54π2\pi r^2 + 2\pi rh = 54\pi, giving h=27−r2rh = \frac{27 - r^2}{r}. Substituting this into the volume equation V=πr2hV = \pi r^2 h gives V(r)=27πr−πr3V(r) = 27\pi r - \pi r^3. Setting the first derivative dVdr=27π−3πr2\frac{dV}{dr} = 27\pi - 3\pi r^2 to zero yields 3πr2=27π3\pi r^2 = 27\pi, so r2=9r^2 = 9 and r=3 cmr = 3\text{ cm}. The second derivative d2Vdr2=−6πr\frac{d^2V}{dr^2} = -6\pi r evaluated at r=3r = 3 is −18π-18\pi, which is strictly negative, confirming that r=3 cmr = 3\text{ cm} maximizes volume.

Step-by-Step Solution

1
Set up the surface area equation for a closed cylinder with the given value.
2πr2+2πrh=54π2\pi r^2 + 2\pi r h = 54\pi
A closed cylinder consists of two circular bases (2πr22\pi r^2) and a curved lateral surface (2πrh2\pi r h).
2
Express hh in terms of rr.
h=27−r2rh = \frac{27 - r^2}{r}
Dividing the surface area equation by 2π2\pi yields r2+rh=27r^2 + rh = 27, allowing hh to be isolated.
3
Substitute hh into the volume formula V=πr2hV = \pi r^2 h to write volume as a function of rr only.
V(r)=27πr−πr3V(r) = 27\pi r - \pi r^3
To maximize volume using calculus, the volume equation must be expressed in terms of a single variable.
4
Differentiate V(r)V(r) with respect to rr and set the derivative equal to zero to find stationary points.
dVdr=27π−3πr2=0  ⟹  r=3\frac{dV}{dr} = 27\pi - 3\pi r^2 = 0 \implies r = 3
Maximum volume occurs at a stationary point where the first derivative is zero.
5
Verify that r=3r = 3 produces a maximum using the second derivative test.
d2Vdr2=−6πr\frac{d^2V}{dr^2} = -6\pi r; at r=3r = 3, d2Vdr2=−18π<0\frac{d^2V}{dr^2} = -18\pi < 0
A negative second derivative indicates a local maximum.

Key Concept

Optimization and Stationary Points in Mensuration
Estimated Time:2m 0s
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Topics

Stationary Points, Maxima, and Minima

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