# Optimization Problems in 3D Geometry

## Solved Problems

Click or tap a problem to see the solution.

### Example 9

A body has the shape of a cylinder, the bases of which are surmounted by hemispheres (Figure 9a). Determine the height of the cylinder H and radius of the hemispheres R, at which the surface area at a fixed volume V is the smallest.

### Example 10

A storage unit has the shape of a hemisphere on top of a cylinder and a fixed volume V (Figure 10a). Determine the dimensions of the solid (the radius R and the height H) that minimize the total surface area.

### Example 11

A cylinder is inscribed in a sphere of radius $$a$$ (Figure $$11a$$). Find the base radius $$R$$ and height $$H$$ of the cylinder with the largest possible volume.

### Example 12

A regular hexagonal prism has the surface area $$S$$ (Figure $$12a$$). What is the largest volume of the prism?

### Example 13

A rectangular box with a square base is inscribed in a hemisphere of radius $$R$$ (Figure $$13a$$). Find the maximum volume of the box.

### Example 14

Suppose that we have a log of length $$H$$ in the form of a frustum of a cone with radii $$R$$ and $$r$$ ($$R \gt r$$). From this log, it is required to cut out a beam with the largest possible volume. The beam has the form of a rectangular parallelepiped with a square cross section and the same axis as the log (Figure $$14a$$).

### Example 15

A square pyramid has a volume $$V.$$ Find the length of the lateral edge that minimizes the total surface area of the pyramid (Figure $$15a$$).

### Example 16

A cone has a volume $$V.$$ What base radius $$R$$ and height $$H$$ will minimize the lateral surface area of the cone (Figure $$16a$$)?

### Example 17

A right tetrahedron with an equilateral triangle in the base is inscribed in a sphere of radius $$R$$ (Figure $$17a$$). Find the largest volume of the tetrahedron.

### Example 9.

A body has the shape of a cylinder, the bases of which are surmounted by hemispheres (Figure $$9a$$). Determine the height of the cylinder $$H$$ and radius of the hemispheres $$R,$$ at which the surface area at a fixed volume $$V$$ is the smallest.

Solution.

The volume of the body is given by

$V = \frac{4}{3}\pi {R^3} + \pi {R^2}H = \pi {R^2}\left( {\frac{{4R}}{3} + H} \right).$

We express here the height $$H:$$

$H = \frac{V}{{\pi {R^3}}} - \frac{{4R}}{3}.$

Consider the total surface area of the body:

$S = 4\pi {R^2} + 2\pi RH.$

Substitute the expression for $$H$$ in the formula above:

$S = S\left( R \right) = 4\pi {R^2} + 2\pi R\left( {\frac{V}{{\pi {R^3}}} - \frac{{4R}}{3}} \right) = 4\pi {R^2} + \frac{{2V}}{R} - \frac{{8\pi {R^2}}}{3} = \frac{{4\pi {R^2}}}{3} + \frac{{2V}}{R}.$

Calculate the derivative of the function $$S\left( R \right):$$

$S'\left( R \right) = \left( {\frac{{4\pi {R^2}}}{3} + \frac{{2V}}{R}} \right)^\prime = \frac{{4\pi }}{3} \cdot 2R - \frac{{2V}}{{{R^2}}} = \frac{{8\pi R}}{3} - \frac{{2V}}{{{R^2}}} = \frac{{8\pi {R^3} - 6V}}{{3{R^2}}}.$

Equating the derivative to zero, we obtain:

$S'\left( R \right) = 0,\;\; \Rightarrow \frac{{8\pi {R^3} - 6V}}{{3{R^2}}} = 0,\;\; \Rightarrow {R^3} = \frac{{3V}}{{4\pi }},\;\; \Rightarrow R = \sqrt[3]{{\frac{{3V}}{{4\pi }}}}.$

The found point, obviously, is the minimum point of the function $$S\left( R \right)$$ as when passing through it the derivative changes sign from minus to plus. Now we calculate the corresponding value of the height of the cylinder:

$H = \frac{V}{{\pi {R^3}}} - \frac{{4R}}{3} = \frac{V}{{\pi {{\left( {\sqrt[3]{{\frac{{3V}}{{4\pi }}}}} \right)}^3}}} - \frac{4}{3}\sqrt[3]{{\frac{{3V}}{{4\pi }}}} = \frac{{V \cdot {4^{\frac{2}{3}}} \cdot {\pi ^{\frac{2}{3}}}}}{{\pi \cdot {3^{\frac{2}{3}}} \cdot {V^{\frac{2}{3}}}}} - \frac{{4 \cdot {3^{\frac{1}{3}}} \cdot {V^{\frac{1}{3}}}}}{{3 \cdot {4^{\frac{1}{3}}} \cdot {\pi ^{\frac{1}{3}}}}} = \frac{{{V^{\frac{1}{3}}} \cdot {4^{\frac{2}{3}}}}}{{{\pi ^{\frac{1}{3}}} \cdot {3^{\frac{2}{3}}}}} - \frac{{{V^{\frac{1}{3}}} \cdot {4^{\frac{2}{3}}}}}{{{\pi ^{\frac{1}{3}}} \cdot {3^{\frac{2}{3}}}}} \equiv 0.$

As seen, the shape of the body should be spherical without a cylindrical part!

### Example 10.

A storage unit has the shape of a hemisphere on top of a cylinder and a fixed volume $$V$$ (Figure $$10a$$). Determine the dimensions of the solid (the radius $$R$$ and the height $$H$$) that minimize the total surface area.

Solution.

The total surface area of the storage unit is given by

$S = \pi {R^2} + 2\pi RH + 2\pi {R^2} = 3\pi {R^2} + 2\pi RH.$

The volume of the solid is

$V = \pi {R^2}H + \frac{2}{3}\pi {R^3}.$

We express the height $$H$$ in terms of $$R$$ and $$V$$ and substitute it into the first equation:

$H = \frac{{V - \frac{2}{3}\pi {R^3}}}{{\pi {R^2}}} = \frac{V}{{\pi {R^2}}} - \frac{2}{3}R.$

Hence

$S = 3\pi {R^2} + 2\pi RH = 3\pi {R^2} + 2\pi R\left( {\frac{V}{{\pi {R^2}}} - \frac{2}{3}R} \right) = 3\pi {R^2} + \frac{{2V}}{R} - \frac{{4\pi {R^2}}}{3} = \frac{{5\pi {R^2}}}{3} + \frac{{2V}}{R} = S\left( R \right).$

Take the derivative:

$S^\prime\left( R \right) = \left( {\frac{{5\pi {R^2}}}{3} + \frac{{2V}}{R}} \right)^\prime = \frac{{10\pi R}}{3} - \frac{{2V}}{{{R^2}}} = \frac{{10\pi {R^3} - 6V}}{{3{R^2}}}.$

Determine the critical value of $$R:$$

$S^\prime\left( R \right) = 0,\;\; \Rightarrow \frac{{10\pi {R^3} - 6V}}{{3{R^2}}}0,\;\; \Rightarrow 10\pi {R^3} - 6V = 0,\;\; \Rightarrow {R^3} = \frac{{3V}}{{5\pi }},\;\; \Rightarrow R = \sqrt[3]{{\frac{{3V}}{{5\pi }}}}.$

Using the First Derivative Test one can show that the point $$R = \sqrt[3]{{\frac{{3V}}{{5\pi }}}}$$ is a point of local minimum.

Calculate the corresponding value of the height $$H:$$

$H = \frac{V}{{\pi {R^2}}} - \frac{2}{3}R = \frac{V}{{\pi {{\left( {\sqrt[3]{{\frac{{3V}}{{5\pi }}}}} \right)}^2}}} - \frac{2}{3}\sqrt[3]{{\frac{{3V}}{{5\pi }}}} = \frac{5}{3} \cdot \frac{{3V}}{{5\pi {{\left( {\sqrt[3]{{\frac{{3V}}{{5\pi }}}}} \right)}^2}}} - \frac{2}{3}\sqrt[3]{{\frac{{3V}}{{5\pi }}}} = \frac{5}{3}\sqrt[3]{{\frac{{3V}}{{5\pi }}}} - \frac{2}{3}\sqrt[3]{{\frac{{3V}}{{5\pi }}}} = \sqrt[3]{{\frac{{3V}}{{5\pi }}}}.$

$R = H = \sqrt[3]{{\frac{{3V}}{{5\pi }}}}.$

### Example 11.

A cylinder is inscribed in a sphere of radius $$a$$ (Figure $$11a$$). Find the base radius $$R$$ and height $$H$$ of the cylinder with the largest possible volume.

Solution.

The volume of the cylinder is

$V = \pi {R^2}H.$

The base radius of the cylinder $$R$$ and the radius of the sphere $$a$$ are connected by the following relationship (Figure $$11a$$):

${a^2} = {R^2} + {\left( {\frac{H}{2}} \right)^2} = {R^2} + \frac{{{H^2}}}{4}.$

Consequently,

${R^2} = {a^2} - \frac{{{H^2}}}{4}.$

Then the expression for the volume of the cylinder can be written as

$V = \pi {R^2}H = \pi \left( {{a^2} - \frac{{{H^2}}}{4}} \right)H = \pi {a^2}H - \frac{{\pi {H^3}}}{4}.$

This relationship is a function $$V\left( H \right),$$ which we will explore further for extreme values. The derivative $$V'\left( H \right)$$ has the form:

$V'\left( H \right) = \left( {\pi {a^2}H - \frac{{\pi {H^3}}}{4}} \right)^\prime = \pi {a^2} - \frac{{3\pi {H^2}}}{4} = \frac{\pi }{4}\left( {4{a^2} - 3{H^2}} \right).$

The roots of the derivative are

$V'\left( H \right) = 0,\;\; \Rightarrow 4{a^2} - 3{H^2} = 0,\;\; \Rightarrow {H^2} = \frac{{4{a^2}}}{3},\;\; \Rightarrow H = \pm \frac{{2a}}{{\sqrt 3 }}.$

Of course, only the positive value $$H = {\frac{{2a}}{{\sqrt 3 }}}$$ is relevant here. When passing through this point the derivative changes sign from plus to minus, that is the function $$V\left( H \right)$$ has a maximum here. The base radius of the cylinder at height $$H$$ is given by

${R^2} = {a^2} - \frac{{{H^2}}}{4} = {a^2} - \frac{1}{4}{\left( {\frac{{2a}}{{\sqrt 3 }}} \right)^2} = {a^2} - \frac{{4{a^2}}}{{12}} = {a^2} - \frac{{{a^2}}}{3} = \frac{{2{a^2}}}{3},\;\; \Rightarrow R = a\sqrt {\frac{2}{3}} .$

So, a cylinder inscribed in a sphere has the largest volume under the following conditions:

$H = \frac{{2a}}{{\sqrt 3 }},\;\;\;R = a\sqrt {\frac{2}{3}} ,$

where $$a$$ is the radius of the sphere. The maximum volume is equal to

$V = \pi {R^2}H = \pi {\left( {a\sqrt {\frac{2}{3}} } \right)^2} \cdot \frac{{2a}}{{\sqrt 3 }} = \frac{{2\pi {a^2}}}{3} \cdot \frac{{2a}}{{\sqrt 3 }} = \frac{{4\pi {a^3}}}{{3\sqrt 3 }},$

that is $$\sqrt 3$$ times less than the volume of the sphere.

### Example 12.

A regular hexagonal prism has the surface area $$S$$ (Figure $$12a$$). What is the largest volume of the prism?

Solution.

Let $$a$$ be the side of the base and $$H$$ be the height of the prism. The area of the base is given by

${A_B} = \frac{{{a^2}\sqrt 3 }}{4} \cdot 6 = \frac{{3\sqrt 3 {a^2}}}{2}.$

Then the surface area of the prism is expressed by the formula

$S = 2{A_B} + {A_L} = 2 \cdot \frac{{3\sqrt 3 {a^2}}}{2} + 6aH = 3\sqrt 3 {a^2} + 6aH.$

We solve the last equation for $$H:$$

$S = 3\sqrt 3 {a^2} + 6aH,\;\; \Rightarrow 6aH = S - 3\sqrt 3 {a^2},\;\; \Rightarrow H = \frac{{S - 3\sqrt 3 {a^2}}}{{6a}} = \frac{S}{{6a}} - \frac{{\sqrt 3 a}}{2}.$

Given that the volume of the prism is

$V = {A_B}H = \frac{{3\sqrt 3 {a^2}}}{2}H,$

we can write it in the form

$V = \frac{{3\sqrt 3 {a^2}}}{2}H = \frac{{3\sqrt 3 {a^2}}}{2}\left( {\frac{S}{{6a}} - \frac{{\sqrt 3 a}}{2}} \right) = \frac{{\sqrt 3 aS - 9{a^3}}}{4} = V\left( a \right).$

Take the derivative and find the critical points:

$V^\prime\left( a \right) = \left( {\frac{{\sqrt 3 aS - 9{a^3}}}{4}} \right)^\prime = \frac{{\sqrt 3 S - 27{a^2}}}{4};$
$V^\prime\left( a \right) = 0,\;\; \Rightarrow \frac{{\sqrt 3 S - 27{a^2}}}{4} = 0,\;\; \Rightarrow \sqrt 3 S - 27{a^2} = 0,\;\; \Rightarrow {a^2} = \frac{{\sqrt 3 S}}{{27}},\;\; \Rightarrow a = \frac{{\sqrt[4]{3}\sqrt S }}{{3\sqrt 3 }} = \frac{{\sqrt S }}{{3\sqrt[4]{3}}}.$

Note that the second derivative is

$V^{\prime\prime}\left( a \right) = - \frac{{54a}}{4} = - \frac{{27a}}{2}.$

It is negative for $$a \gt 0,$$ so the point $$a = \frac{{\sqrt S }}{{3\sqrt[4]{3}}}$$ is a point of local maximum by the Second Derivative Test.

Now we can calculate the maximum volume of the prism:

${V_{\max }} = V\left( {a = \frac{{\sqrt S }}{{3\sqrt[4]{3}}}} \right) = \frac{{\sqrt 3 \cdot \frac{{\sqrt S }}{{3\sqrt[4]{3}}} \cdot S - 9{{\left( {\frac{{\sqrt S }}{{3\sqrt[4]{3}}}} \right)}^3}}}{4} = \frac{{{3^{\frac{1}{2}}} \cdot {S^{\frac{3}{2}}}}}{{3 \cdot {3^{\frac{1}{4}}}}} - \frac{{{3^2} \cdot {S^{\frac{3}{2}}}}}{{{3^3} \cdot {3^{\frac{3}{4}}}}} = \frac{{{S^{\frac{3}{2}}}}}{{{3^{\frac{3}{4}}}}} - \frac{{{S^{\frac{3}{2}}}}}{{{3^{\frac{7}{4}}}}} = \frac{{{S^{\frac{3}{2}}}}}{{{3^{\frac{3}{4}}}}}\left( {1 - \frac{1}{3}} \right) = \frac{2}{3} \cdot \frac{{{S^{\frac{3}{2}}}}}{{{3^{\frac{3}{4}}}}} = \frac{{2\sqrt {{S^3}} }}{{3\sqrt[4]{3}}}.$

### Example 13.

A rectangular box with a square base is inscribed in a hemisphere of radius $$R$$ (Figure $$13a$$). Find the maximum volume of the box.

Solution.

The base of the rectangular box lies in the plane that contains the base of the hemisphere.

Using the Pythagorean theorem, we can write the relationship:

${\left( {\sqrt 2 \cdot \frac{x}{2}} \right)^2} + {y^2} = {R^2},\;\; \Rightarrow \frac{{{x^2}}}{2} + {y^2} = {R^2}.$

Hence

$y = \sqrt {{R^2} - \frac{{{x^2}}}{2}} .$

The volume of the inscribed box is given by

$V = {x^2}y = {x^2}\sqrt {{R^2} - \frac{{{x^2}}}{2}} = \frac{{{x^2}}}{{\sqrt 2 }}\sqrt {2{R^2} - {x^2}} = V\left( x \right).$

The derivative of the function $$V\left( x \right)$$ is written in the form

$V^\prime\left( x \right) = \left( {\frac{{{x^2}}}{{\sqrt 2 }}\sqrt {2{R^2} - {x^2}} } \right)^\prime = \frac{1}{{\sqrt 2 }} \cdot \frac{{4{R^2}x - 3{x^3}}}{{\sqrt {2{R^2} - {x^2}} }}.$

Using the First Derivative Test, we find that the function $$V\left( x \right)$$ has a maximum at $$x = \frac{{2R}}{{\sqrt 3 }}.$$

Calculate the value of $$y:$$

$y = \sqrt {{R^2} - \frac{{{x^2}}}{2}} = \sqrt {{R^2} - \frac{{{{\left( {\frac{{2R}}{{\sqrt 3 }}} \right)}^2}}}{2}} = \sqrt {{R^2} - \frac{{2{R^2}}}{3}} = \sqrt {\frac{{{R^2}}}{3}} = \frac{R}{{\sqrt 3 }}.$

Then the maximum volume of the box is equal to

${V_{\max }} = {x^2}y = \frac{{{{\left( {\frac{{2R}}{{\sqrt 3 }}} \right)}^2}}}{2} \cdot \frac{R}{{\sqrt 3 }} = \frac{{8{R^2}}}{3} \cdot \frac{R}{{\sqrt 3 }} = \frac{{8{R^3}}}{{3\sqrt 3 }}.$

### Example 14.

Suppose that we have a log of length $$H$$ in the form of a frustum of a cone with radii $$R$$ and $$r$$ ($$R \gt r$$). From this log, it is required to cut out a beam with the largest possible volume. The beam has the form of a rectangular parallelepiped with a square cross section and the same axis as the log (Figure $$14a$$).

Solution.

The frustum of cone and the inscribed parallelepiped are schematically shown in cross section in Figure $$14a.$$

The volume of the parallelepiped is defined by the formula

$V = {x^2}y,$

where $$x$$ is the side of the square base, $$y$$ is the height of the parallelepiped.

Considering the similar triangles $$CBR$$ and $$CKM,$$ we can write the following proportion:

$\frac{{MK}}{{AB}} = \frac{{MC}}{{AC}},\;\;\Rightarrow \frac{y}{H} = \frac{{R - \frac{x}{2}}}{{R - r}}.$

Hence, the height $$y$$ is given by

$y = \frac{{H\left( {R - \frac{x}{2}} \right)}}{{R - r}}.$

Write the volume $$V$$ as a function $$x:$$

$V = V\left( x \right) = {x^2}y = \frac{{{x^2}H\left( {R - \frac{x}{2}} \right)}}{{R - r}} = \frac{H}{{R - r}}\left( {R{x^2} - \frac{{{x^3}}}{2}} \right).$

The derivative is written as

$V'\left( x \right) = {\left[ {\frac{H}{{R - r}}\left( {R{x^2} - \frac{{{x^3}}}{2}} \right)} \right]^\prime } = \frac{H}{{R - r}}\left( {2Rx - \frac{{3{x^2}}}{2}} \right) = \frac{{Hx}}{{R - r}}\left( {2R - \frac{{3x}}{2}} \right).$

Find the stationary point:

$V'\left( x \right) = 0,\;\; \Rightarrow \frac{{Hx}}{{R - r}}\left( {2R - \frac{{3x}}{2}} \right) = 0,\;\; \Rightarrow 2R - \frac{{3x}}{2} = 0,\;\; \Rightarrow x = \frac{{4R}}{3}.$

The derivative is positive to the left of this point and negative to the right. Hence, this is a point of maximum of the function $$V\left( x \right).$$ Then the height of the parallelepiped is

$y = \frac{{H\left( {R - \frac{x}{2}} \right)}}{{R - r}} = \frac{{H\left( {R - \frac{{4R}}{6}} \right)}}{{R - r}} = \frac{{HR\left( {1 - \frac{2}{3}} \right)}}{{R - r}} = \frac{{HR}}{{3\left( {R - r} \right)}}.$

So, the parallelepiped inscribed in the truncated cone has the largest volume when its sides are equal

$x = \frac{{4R}}{3},\;\;\;y = \frac{{HR}}{{3\left( {R - r} \right)}}.$

### Example 15.

A square pyramid has a volume $$V.$$ Find the length of the lateral edge that minimizes the total surface area of the pyramid (Figure $$15a$$).

Solution.

Let $$a$$ be the side of the square base and $$H$$ be the height of the pyramid. The lateral edge is denoted by $$b.$$

The slant height $$s$$ of the pyramid can be found using Pythagorean Theorem:

${s^2} = {H^2} + {\left( {\frac{a}{2}} \right)^2} = {H^2} + \frac{{{a^2}}}{4}.$

The lateral surface area of the pyramid is written as

${A_L} = 4 \cdot \frac{1}{2}as = 2a\sqrt {{H^2} + \frac{{{a^2}}}{4}} = a\sqrt {4{H^2} + {a^2}} .$

Then the total surface area is expressed in the form

$S = {A_L} + {A_B} = a\sqrt {4{H^2} + {a^2}} + {a^2}.$

The volume of the pyramid is given by

$V = \frac{1}{3}{a^2}H,$

so that

$H = \frac{{3V}}{{{a^2}}}.$

We substitute the expression for $$H$$ in the formula for the surface area:

$S = a\sqrt {4{H^2} + {a^2}} + {a^2} = a\sqrt {4 \cdot {{\left( {\frac{{3V}}{{{a^2}}}} \right)}^2} + {a^2}} + {a^2} = a\sqrt {\frac{{36{V^2}}}{{{a^4}}} + {a^2}} + {a^2} = \frac{{\sqrt {36{V^2} + {a^6}} }}{a} + {a^2} = S\left( a \right).$

Since the surface area $$S$$ is written now as a function $$S\left( a \right),$$ we can examine its extreme values. The derivative is given by (we omit the details here):

$S^\prime\left( a \right) = \left( {\frac{{\sqrt {36{V^2} + {a^6}} }}{a} + {a^2}} \right)^\prime = \frac{{2{a^6} - 36{V^2}}}{{{a^2}\sqrt {36{V^2} + {a^6}} }} + 2a.$

Determine the critical point by equating the derivative to zero:

$S^\prime\left( a \right) = 0,\;\; \Rightarrow \frac{{2{a^6} - 36{V^2}}}{{{a^2}\sqrt {36{V^2} + {a^6}} }} + 2a = 0.$

Solving the last equation, we get the positive critical value of $$a:$$

$a = \sqrt[6]{{\frac{9}{2}{V^2}}} = \frac{{\sqrt[3]{{3V}}}}{{\sqrt[6]{2}}}.$

Using the First Derivative Test, one can show that this is a point of local minimum which corresponds to the smallest surface area of the pyramid.

Find the height of the pyramid:

$H = \frac{{3V}}{{{a^2}}} = \frac{{3V}}{{{{\left( {\frac{{\sqrt[3]{{3V}}}}{{\sqrt[6]{2}}}} \right)}^2}}} = \frac{{3V \cdot \sqrt[3]{2}}}{{\sqrt[3]{{{{\left( {3V} \right)}^2}}}}} = \sqrt[3]{{3V}} \cdot \sqrt[3]{2} = \sqrt[3]{{6V}}.$

Now we can calculate the length of the lateral edge of the optimal pyramid:

$b = \sqrt {{{\left( {\frac{{\sqrt 2 }}{2}a} \right)}^2} + {H^2}} = \sqrt {\frac{{{a^2}}}{2} + {H^2}} = \sqrt {\frac{{{{\left( {\frac{{\sqrt[3]{{3V}}}}{{\sqrt[6]{2}}}} \right)}^2}}}{2} + {{\left( {\sqrt[3]{{6V}}} \right)}^2}} = \sqrt {\frac{{\sqrt[3]{{9{V^2}}}}}{{2\sqrt[3]{2}}} + \sqrt[3]{{36{V^2}}}} = \sqrt {\frac{{\sqrt[3]{{9{V^2}}} + \sqrt[3]{{16}} \cdot \sqrt[3]{{36{V^2}}}}}{{\sqrt[3]{{16}}}}} = \sqrt {\frac{{\sqrt[3]{{9{V^2}}} + \sqrt[3]{{576{V^2}}}}}{{\sqrt[3]{{16}}}}} = \sqrt {\frac{{\sqrt[3]{{9{V^2}}} + 4\sqrt[3]{{9{V^2}}}}}{{\sqrt[3]{{16}}}}} = \sqrt {\frac{{5\sqrt[3]{{9{V^2}}}}}{{\sqrt[3]{{16}}}}} = \frac{{\sqrt 5 \sqrt[6]{{9{V^2}}}}}{{\sqrt[3]{4}}} = \frac{{\sqrt 5 \sqrt[3]{{3V}}}}{{\sqrt[6]{2}}}.$

### Example 16.

A cone has a volume $$V.$$ What base radius $$R$$ and height $$H$$ will minimize the lateral surface area of the cone (Figure $$16a$$)?

Solution.

Let the slant height of the cone be denoted by $$m$$ (Figure $$16a$$).

The lateral surface area is expressed by the formula

${S_{\text{L}}} = \pi Rm.$

Further we will denote the lateral surface area of the cone simply by the letter $$S.$$ Given that the volume of the cone is equal to

$V = \frac{1}{3}\pi {R^2}H,$

we express the height $$H$$ in terms of $$R$$ and $$V:$$

$H = \frac{{3V}}{{\pi {R^2}}}.$

By the Pythagorean theorem, we find:

$m = \sqrt {{H^2} + {R^2}} = \sqrt {{{\left( {\frac{{3V}}{{\pi {R^2}}}} \right)}^2} + {R^2}} .$

Then the lateral surface area is written as a function of the base radius $$R:$$

$S = \pi Rm = \pi R\sqrt {{{\left( {\frac{{3V}}{{\pi {R^2}}}} \right)}^2} + {R^2}} = \pi R\sqrt {\frac{{9{V^2}}}{{{\pi ^2}{R^4}}} + {R^2}} = \pi R\sqrt {\frac{{9{V^2} + {\pi ^2}{R^6}}}{{{\pi ^2}{R^4}}}} = \frac{{\cancel{\pi} \cancel{R}}}{{\cancel{\pi} {R^{\cancel{2}}}}}\sqrt {9{V^2} + {\pi ^2}{R^6}} = \frac{{\sqrt {9{V^2} + {\pi ^2}{R^6}} }}{R}.$

Compute the derivative

$S'\left( R \right) = {\left( {\frac{{\sqrt {9{V^2} + {\pi ^2}{R^6}} }}{R}} \right)^\prime } = \frac{{\frac{{6{\pi ^2}{R^5}}}{{2\sqrt {9{V^2} + {\pi ^2}{R^6}} }} \cdot R - \sqrt {9{V^2} + {\pi ^2}{R^6}}}}{{{R^2}}} = \frac{{6{\pi ^2}{R^6} - 2\left( {9{V^2} + {\pi ^2}{R^6}} \right)}}{{2{R^2}\sqrt {9{V^2} + {\pi ^2}{R^6}} }} = \frac{{4{\pi ^2}{R^6} - 18{V^2}}}{{2{R^2}\sqrt {9{V^2} + {\pi ^2}{R^6}} }} = \frac{{2{\pi ^2}{R^6} - 9{V^2}}}{{{R^2}\sqrt {9{V^2} + {\pi ^2}{R^6}} }}.$

The derivative is zero provided

$2{\pi ^2}{R^6} - 9{V^2} = 0,\;\; \Rightarrow {R^6} = \frac{{9{V^2}}}{{2{\pi ^2}}},\;\; \Rightarrow R = \sqrt[6]{{\frac{{9{V^2}}}{{2{\pi ^2}}}}}.$

It is seen that with increasing $$R$$ and passing through the above critical point the derivative changes sign from minus to plus. Consequently, the function $$S\left( R \right)$$ has a minimum at this point.

Determine the height of the cone:

$H = \frac{{3V}}{{\pi {R^2}}} = \frac{{3V}}{{\pi \sqrt[3]{{\frac{{9{V^2}}}{{2{\pi ^2}}}}}}} = \frac{{3V}}{{\pi \cdot \frac{{{3^{\frac{2}{3}}} \cdot {V^{\frac{2}{3}}}}}{{{2^{\frac{1}{3}}} \cdot {\pi ^{\frac{2}{3}}}}}}} = \frac{{{3^{\frac{1}{3}}} \cdot {V^{\frac{1}{3}}} \cdot {2^{\frac{1}{3}}}}}{{{\pi ^{\frac{1}{3}}}}} = \sqrt[3]{{\frac{{6V}}{\pi }}}.$

To better understand the optimal shape of the cone, we calculate the ratio $$\frac{H}{R}:$$

$\frac{H}{R} = \frac{{\sqrt[3]{{\frac{{6V}}{\pi }}}}}{{\sqrt[6]{{\frac{{9{V^2}}}{{2{\pi ^2}}}}}}} = \frac{{{3^{\frac{1}{3}}} \cdot {2^{\frac{1}{3}}} \cdot {V^{\frac{1}{3}}}}}{{{\pi ^{\frac{1}{3}}}}}:\frac{{{3^{\frac{1}{3}}} \cdot {V^{\frac{1}{3}}}}}{{{2^{\frac{1}{6}}} \cdot {\pi ^{\frac{1}{3}}}}} = \frac{{\cancel{3^{\frac{1}{3}}} \cdot {2^{\frac{1}{3}}} \cdot \cancel{V^{\frac{1}{3}}} \cdot \cancel{\pi ^{\frac{1}{3}}} \cdot {2^{\frac{1}{6}}}}}{{\cancel{\pi ^{\frac{1}{3}}} \cdot \cancel{3^{\frac{1}{3}}} \cdot \cancel{V^{\frac{1}{3}}}}} = 2^{\frac{1}{2}} = \sqrt 2 .$

Thus, the height of the cone with the least lateral surface area should be about $$1,4$$ times greater than the base radius.

It is interesting what is the ratio of the height to the base radius of such cone-shaped mega construction project as Khan Shatyr Entertainment Center (Figure $$16b$$)? If the lateral surface area was one of the critical factors in the design, it is likely that its shape should be close to our solution.

### Example 17.

A right tetrahedron with an equilateral triangle in the base is inscribed in a sphere of radius $$R$$ (Figure $$17a$$). Find the largest volume of the tetrahedron.

Solution.

Let $$a$$ be the side of tetrahedron base and $$H$$ be the height of the tetrahedron.

The triangle $$\triangle ABC$$ is isosceles, the angle $$\angle MBC = 30^\circ.$$ Then

$\left| {MB} \right| = \frac{{\frac{a}{2}}}{{\cos 30^\circ}} = \frac{a}{{2 \cdot \frac{{\sqrt 3 }}{2}}} = \frac{a}{{\sqrt 3 }}.$

Using the Pythagorean Theorem, we can write the constraint equation:

${\left| {OM} \right|^2} + {\left| {MB} \right|^2} = {R^2},\;\; \Rightarrow {\left( {H - R} \right)^2} + {\left( {\frac{a}{{\sqrt 3 }}} \right)^2} = {R^2},\;\; \Rightarrow {H^2} - 2HR + {R^2} + \frac{{{a^2}}}{3} = {R^2},\;\; \Rightarrow {H^2} - 2HR + \frac{{{a^2}}}{3} = 0,\;\; \Rightarrow {a^2} = 6HR - 3{H^2}.$

The volume of the tetrahedron is written as

$V = \frac{1}{3} \cdot \frac{{{a^2}\sqrt 3 }}{4} \cdot H = \frac{{\sqrt 3 }}{{12}}\left( {6HR - 3{H^2}} \right)H = \frac{{\sqrt 3 }}{{12}}\left( {6{H^2}R - 3{H^3}} \right) = \frac{{\sqrt 3 }}{4}\left( {2{H^2}R - {H^3}} \right) = V\left( H \right).$

Take the derivative:

$V^\prime\left( H \right) = \left[ {\frac{{\sqrt 3 }}{4}\left( {2{H^2}R - {H^3}} \right)} \right]^\prime = \frac{{\sqrt 3 }}{4}\left( {4HR - 3{H^2}} \right).$

Find the critical points:

$V^\prime\left( H \right) = 0,\;\; \Rightarrow \frac{{\sqrt 3 }}{4}\left( {4HR - 3{H^2}} \right) = 0,\;\; \Rightarrow H\left( {4R - 3H} \right) = 0,\;\; \Rightarrow H = 0,\,\frac{{4R}}{3}.$

Hence, the positive critical value is equal $$H = \frac{{4R}}{3}.$$

Compute the second derivative:

$V^{\prime\prime}\left( H \right) = \left[ {\frac{{\sqrt 3 }}{4}\left( {4HR - 3{H^2}} \right)} \right]^\prime = \frac{{\sqrt 3 }}{4}\left( {4R - 6H} \right).$

We see that

$V^{\prime\prime}\left( {H = \frac{{4R}}{3}} \right) = \frac{{\sqrt 3 }}{4}\left( {4R - 6 \cdot \frac{{4R}}{3}} \right) = \frac{{\sqrt 3 }}{4} \cdot \left( { - 4R} \right) = - \sqrt 3 R \lt 0,$

that is the point $$H = \frac{{4R}}{3}$$ is a point of local maximum by the Second Derivative Test.

Calculate the value of $${{a^2}}:$$

${a^2} = 6HR - 3{H^2} = 6 \cdot \frac{{4R}}{3} \cdot R - 3 \cdot {\left( {\frac{{4R}}{3}} \right)^2} = 8{R^2} - 3 \cdot \frac{{16{R^2}}}{9} = 8{R^2} - \frac{{16{R^2}}}{3} = \frac{{8{R^2}}}{3}.$

Then the maximum possible volume of the inscribed tetrahedron is given by

${V_{\max }} = \frac{1}{3} \cdot \frac{{\sqrt 3 {a^2}}}{4} \cdot H = \frac{{\sqrt 3 }}{{12}} \cdot \frac{{8{R^2}}}{3} \cdot \frac{{4R}}{3} = \frac{{8\sqrt 3 {R^3}}}{{27}}.$