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Here is the problem:

A margarine tub has the shape of the frustum of a cone. With the lower base having diameter length 11 cm and the upper base having diameter length 14 cm, the volume of such a container 6 ⅔ cm tall can be determined by using H = 32 ⅔ cm, and h = 26 cm. Find its volume.

Let

R = ¹⁴⁄₂ = 7 r = ¹¹⁄₂ = 5.5 H = 32 ⅔ cm h = 26 cm k = H - h = 6 ⅔

So the solution in the book is using the formula

V = ⅓πR²H - ⅓πr²h

Which is derived from this image: enter image description here

doing so we have

V = ⅓ π (7 cm)² (32 ⅔ cm) - ⅓ π (5.5 cm)² (26 cm)

V ≈ 852.593 cm³

To check, I used the other formula:

V = ⅓πk (R² + Rr + r²) ; wherein k = H - h

and the result is

V ≈ 822.050 cm³

Why there are different answers, where does the error came from?

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  • $\begingroup$ "Frustum", not "frustRum", by the way. $\endgroup$ Commented Jul 8 at 21:15
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    $\begingroup$ Indeed, the word is frustratingly easy to misspell. $\endgroup$ Commented Jul 9 at 0:01

1 Answer 1

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The difference comes from the inconsistent information in the question:

  • $\dfrac{d}{D}=\dfrac{11}{14}\approx 0.7857$

  • $\dfrac{h}{H}=\dfrac{26}{32+\frac23}\approx 0.7723$

If instead you used $h=\frac{11}{3}(6+\frac23)=24+\frac49$ and $H=\frac{14}{3}(6+\frac23)=31+\frac19$ then you would again get $V \approx 822.050$, and consistent answers.

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  • $\begingroup$ Does it mean that the cone doesn't "converge" where it should? $\endgroup$ Commented Jul 9 at 8:13
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    $\begingroup$ @EricDuminil It does not "converge" where the question suggests. $\endgroup$
    – Henry
    Commented Jul 9 at 8:59

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