Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius 'a' is $\dfrac{8}{27}$ (volume of the sphere). **OR** With usual notations, show that $(Z_n, +_n)$ forms a group.

> [!TLDR] > Optimizing the volume of a cone inscribed in a sphere of radius $a$ shows the maximum volume is $\dfrac{8}{27}$ of the sphere's volume; the OR alternative verifies the four group axioms for $(\mathbb{Z}_n,+_n)$. > > **Main part** > > 1. Let the sphere have centre $O$ and radius $a$. Inscribe a right circular cone with its apex $A$ on the sphere and its axis along a diameter; let the cone have height $h$ (measured from apex to the base plane) and base radius $r$. > 2. If the base circle lies on the sphere, its centre is at signed axial distance $(h-a)$ from $O$ (taking the apex as origin of the axis, sphere centre at distance $a$ from the apex). By the sphere's equation, the base-circle radius satisfies $r^2 = a^2-(a-h)^2 = a^2-a^2+2ah-h^2 = 2ah-h^2 = h(2a-h)$, valid for $0 3. Volume of the cone: $V = \dfrac13\pi r^2h = \dfrac13\pi h(2a-h)h = \dfrac13\pi(2ah^2-h^3)$. > 4. Differentiate with respect to $h$: $\dfrac{dV}{dh} = \dfrac13\pi(4ah-3h^2) = \dfrac13\pi h(4a-3h)$. > 5. Set $\dfrac{dV}{dh}=0$: since $0 6. Second derivative: $\dfrac{d^2V}{dh^2} = \dfrac13\pi(4a-6h)$. At $h=\dfrac{4a}{3}$: $4a-6\left(\dfrac{4a}{3}\right) = 4a-8a=-4a 7. At $h=\dfrac{4a}{3}$: $r^2 = h(2a-h) = \dfrac{4a}{3}\left(2a-\dfrac{4a}{3}\right) = \dfrac{4a}{3}\cdot\dfrac{2a}{3} = \dfrac{8a^2}{9}$. > 8. Maximum volume: $V_{max} = \dfrac13\pi r^2h = \dfrac13\pi\cdot\dfrac{8a^2}{9}\cdot\dfrac{4a}{3} = \dfrac{32}{81}\pi a^3$. > 9. Volume of the sphere: $V_{sphere} = \dfrac43\pi a^3$. > 10. Ratio: $\dfrac{V_{max}}{V_{sphere}} = \dfrac{32\pi a^3/81}{4\pi a^3/3} = \dfrac{32}{81}\times\dfrac{3}{4} = \dfrac{96}{324} = \dfrac{8}{27}$. > 11. Hence $V_{max} = \dfrac{8}{27}\,V_{sphere}$, i.e. the largest cone inscribable in the sphere has $\dfrac8{27}$ of the sphere's volume — as required. > > **OR — alternative** > > 1. Let $\mathbb Z_n=\{0,1,2,\dots,n-1\}$ with the binary operation $+_n$ defined

Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius 'a' is (8)

Question 109 of 148

\dfrac{8}{27}

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Concept understanding — Maxima and Minima

Absolute (global) extrema. For $f$ defined on a domain $D$ , $f(x_0)$ is the absolute maximum of $f$ on $D$ if $f(x_0)\ge f(x)$ for every $x\in D$ ; the absolute minimum is defined symmetrically with $\le$ .

Extreme Value Theorem. If $f$ is continuous on a closed interval $[a,b]$ , then $f$ attains both an absolute maximum and an absolute minimum somewhere on $[a,b]$ — and the extremum can only occur either at an interior critical number or at one of the two endpoints.

Procedure for absolute extrema on $[a,b]$ (Exercise 7.6 Q1's method):

Find every critical number of $f$ in the open interval $(a,b)$ .
Evaluate $f$ at each critical number and at both endpoints $a,b$ .
The largest of these values is the absolute maximum; the smallest is the absolute minimum.

Relative (local) extrema. $f$ has a relative (local) maximum at $x_0$ if $f(x_0)$ is the largest value of $f$ on some open interval around $x_0$ (relative minimum: smallest, on some open interval). A function may have several local extrema, and a local extremum need not be the absolute one.

Fermat's Theorem. If $f$ has a relative extremum at $x=c$ , then $c$ must be a critical number of $f$ (so the search for local extrema always starts by solving $f'(x)=0$ together with any points where $f'$ fails to exist) — though not every critical number is automatically an extremum (e.g. $y=x^3$ at $x=0$ ).

First Derivative Test. At a critical point $c$ where $f$ is continuous, examine the sign of $f'(x)$ moving left to right across $c$ :

negative $\to$ positive: local minimum at $c$ ;
positive $\to$ negative: local maximum at $c$ ;
no sign change (same sign on both sides): $c$ is neither a local max nor a local min.

Second Derivative Test (an alternative, often quicker, at a stationary point). If $f'(c)=0$ and $f''(c)$ exists:

$f''(c)<0$ $\Rightarrow$ local maximum at $c$ ;
$f''(c)>0$ $\Rightarrow$ local minimum at $c$ ; …

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Optimizing the volume of a cone inscribed in a sphere of radius $a$ shows the maximum volume is $\dfrac{8}{27}$ of the sphere's volume; the OR alternative verifies the four group axioms for $(\mathbb{Z}_n,+_n)$ .

Main part

Let the sphere have centre $O$ and radius $a$ . Inscribe a right circular cone with its apex $A$ on the sphere and its axis along a diameter; let the cone have height $h$ (measured from apex to the base plane) and base radius $r$ .
If the base circle lies on the sphere, its centre is at signed axial distance $(h-a)$ from $O$ (taking the apex as origin of the axis, sphere centre at distance $a$ from the apex). By the sphere's equation, the base-circle radius satisfies $r^2 = a^2-(a-h)^2 = a^2-a^2+2ah-h^2 = 2ah-h^2 = h(2a-h)$ , valid for $0<h<2a$ .
Volume of the cone: $V = \dfrac13\pi r^2h = \dfrac13\pi h(2a-h)h = \dfrac13\pi(2ah^2-h^3)$ .
Differentiate with respect to $h$ : $\dfrac{dV}{dh} = \dfrac13\pi(4ah-3h^2) = \dfrac13\pi h(4a-3h)$ .
Set $\dfrac{dV}{dh}=0$ : since $0<h<2a$ , $h\ne0$ , so $4a-3h=0 \Rightarrow h=\dfrac{4a}{3}$ .
Second derivative: $\dfrac{d^2V}{dh^2} = \dfrac13\pi(4a-6h)$ . At $h=\dfrac{4a}{3}$ : $4a-6\left(\dfrac{4a}{3}\right) = 4a-8a=-4a<0$ , confirming a maximum.
At $h=\dfrac{4a}{3}$ : $r^2 = h(2a-h) = \dfrac{4a}{3}\left(2a-\dfrac{4a}{3}\right) = \dfrac{4a}{3}\cdot\dfrac{2a}{3} = \dfrac{8a^2}{9}$ .
Maximum volume: $V_{max} = \dfrac13\pi r^2h = \dfrac13\pi\cdot\dfrac{8a^2}{9}\cdot\dfrac{4a}{3} = \dfrac{32}{81}\pi a^3$ .
Volume of the sphere: $V_{sphere} = \dfrac43\pi a^3$ .
Ratio: $\dfrac{V_{max}}{V_{sphere}} = \dfrac{32\pi a^3/81}{4\pi a^3/3} = \dfrac{32}{81}\times\dfrac{3}{4} = \dfrac{96}{324} = \dfrac{8}{27}$ .
Hence $V_{max} = \dfrac{8}{27}\,V_{sphere}$ , i.e. the largest cone inscribable in the sphere has $\dfrac8{27}$ of the sphere's volume — as required.

OR — alternative

Let $\mathbb Z_n=\{0,1,2,\dots,n-1\}$ with the binary operation $+_n$ defined by $a+_n b = (a+b)\bmod n$ (addition modulo $n$ ).
Closure: for any $a,b\in\mathbb Z_n$ , $(a+b)\bmod n$ is one of $0,1,\dots,n-1$ , so $a+_nb\in\mathbb Z_n$ . Closure holds.
Associativity: for $a,b,c\in\mathbb Z_n$ , $(a+_nb)+_nc = \big[(a+b)\bmod n + c\big]\bmod n = (a+b+c)\bmod n$ , since reducing modulo $n$ partway through an integer sum does not change the final result mod $n$ . Likewise $a+_n(b+_nc) = \big[a+(b+c)\bmod n\big]\bmod n = (a+b+c)\bmod n$ . Both equal $(a+b+c)\bmod n$ , so $(a+_nb)+_nc = a+_n(b+_nc)$ ; associativity holds. …

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Q.Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius 'a' is 827\dfrac{8}{27}278​ (volume of the sphere). OR With usual notations, show that (Zn,+n)(Z_n, +_n)(Zn​,+n​) forms a group.

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Q.Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius 'a' is $\dfrac{8}{27}$ (volume of the sphere). OR With usual notations, show that $(Z_n, +_n)$ forms a group.

Unlock everything free for 14 days