This topic covers different optimization problems related to basic solid shapes (Pyramid, Cone, Cylinder, Prism, Sphere).

To solve such problems you can use the general approach discussed on the page Optimization Problems in 2D Geometry.

## Solved Problems

Click a problem to see the solution.

### Example 1

A sphere of radius \(r\) is inscribed in a right circular cone (Figure \(1a\)). Find the minimum volume of the cone.### Example 2

Find the cylinder with the smallest surface area (Figure \(2a\)).### Example 3

Given a cone with a slant height \(l\) (Figure \(3a\)). Find the largest possible volume of the cone.### Example 4

Determine the largest volume of a cylinder inscribed in a cone with height \(H\) and base radius \(R\) (Figure \(4a\)).### Example 5

A coffee filter has a conical shape (Figure \(5a\)). It is made from a circle sector with the central angle \(\alpha\) such that the volume of coffee held in the filter is maximized. Determine the central angle \(\alpha.\)### Example 6

Find a cone with the largest volume inscribed in a sphere of radius \(R\) (Figure \(6a\)).### Example 7

A piece of cardboard is a rectangle of sides \(a\) and \(b.\) An equal square is cut out from each corner and the sides are folded up to make an open-top rectangular box (Figure \(7a\)). How large the square should be to make the box with the largest possible volume?### Example 8

A Chinese tea bowl has the shape of a spherical cap with the radius \(r\) and the height \(h\) (Figure \(8a\)). Determine the ratio \(\large{\frac{h}{r}}\normalsize\) that maximizes the volume of the bowl for a fixed surface area.### 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 1.

A sphere of radius \(r\) is inscribed in a right circular cone (Figure \(1a\)). Find the minimum volume of the cone.Solution.

The volume of a cone is given by the formula

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

where \(R\) is the radius of the base, \(H\) is the height.

The triangles \(\triangle CMO\) and \(\triangle CKB\) are similar. Then

\[{\frac{r}{H}} = {\frac{R}{m}},\]

where \(r\) is the radius of inscribed circle, \(m\) is the slant height of the cone.

Using the Pythagorean theorem, we have

\[{m = \sqrt {{R^2} + {H^2}}} .\]

Hence

\[{\frac{r}{H} = \frac{R}{{\sqrt {{R^2} + {H^2}} }},}\;\; \Rightarrow {RH = r\sqrt {{R^2} + {H^2}} ,}\;\; \Rightarrow {{R^2}{H^2} = {r^2}{H^2} + {r^2}{R^2},}\;\; \Rightarrow {{H^2} = \frac{{{r^2}{R^2}}}{{{R^2} – {r^2}}},}\;\; \Rightarrow {H = \frac{{rR}}{{\sqrt {{R^2} – {r^2}} }}.}\]

Now we can write the volume of the cone as a function of the base radius \(R:\)

\[{V = \frac{1}{3}\pi {R^2}H }={ \frac{1}{3}\pi {R^2} \cdot \frac{{rR}}{{\sqrt {{R^2} – {r^2}} }} }={ \frac{\pi }{3}\frac{{r{R^3}}}{{\sqrt {{R^2} – {r^2}} }} }={ V\left( R \right).}\]

Find the derivative \(V^\prime\left( R \right):\)

\[{V^\prime\left( R \right) }={ \left( {\frac{\pi }{3}\frac{{r{R^3}}}{{\sqrt {{R^2} – {r^2}} }}} \right)^\prime }={ \frac{{\pi r}}{3} \cdot \frac{{3{R^2} \cdot \sqrt {{R^2} – {r^2}} – {R^3} \cdot \frac{{2R}}{{2\sqrt {{R^2} – {r^2}} }}}}{{{{\left( {\sqrt {{R^2} – {r^2}} } \right)}^2}}} }={ \frac{{\pi r}}{3} \cdot \frac{{3{R^2}\left( {{R^2} – {r^2}} \right) – {R^4}}}{{\sqrt {{{\left( {{R^2} – {r^2}} \right)}^3}} }} }={ \frac{{\pi r}}{3} \cdot \frac{{3{R^4} – 3{R^2}{r^2} – {R^4}}}{{\sqrt {{{\left( {{R^2} – {r^2}} \right)}^3}} }} }={ \frac{{\pi r\left( {2{R^4} – 3{R^2}{r^2}} \right)}}{{3\sqrt {{{\left( {{R^2} – {r^2}} \right)}^3}} }} }={ \frac{{\pi r{R^2}\left( {2{R^2} – 3{r^2}} \right)}}{{3\sqrt {{{\left( {{R^2} – {r^2}} \right)}^3}} }}.}\]

Set the derivative equal to zero to calculate the critical point:

\[{V^\prime\left( R \right) = 0,}\;\; \Rightarrow {\frac{{\pi r{R^2}\left( {2{R^2} – 3{r^2}} \right)}}{{3\sqrt {{{\left( {{R^2} – {r^2}} \right)}^3}} }} = 0,\;}\; \Rightarrow {2{R^2} – 3{r^2} = 0,}\;\; \Rightarrow {{R^2} = \frac{{3{r^2}}}{2},}\;\; \Rightarrow {R = \sqrt {\frac{3}{2}} r.}\]

This point is a local minimum by the First Derivative Test.

The height of the cone is

\[{H = \frac{{rR}}{{\sqrt {{R^2} – {r^2}} }} }={ \frac{{r\sqrt {\frac{3}{2}} r}}{{\sqrt {{{\left( {\sqrt {\frac{3}{2}} r} \right)}^2} – {r^2}} }} }={ \frac{{\sqrt {\frac{3}{2}} {r^2}}}{{\sqrt {\frac{3}{2}{r^2} – {r^2}} }} }={ \frac{{\sqrt {\frac{3}{2}} {r^2}}}{{\sqrt {\frac{1}{2}} r}} }={ \sqrt 3 r.}\]

Hence, the minimum volume of the cone is given by

\[{{V_{\min }} = \frac{1}{3}\pi {R^2}H }={ \frac{1}{3}\pi \cdot \frac{3}{2}{r^2} \cdot \sqrt 3 r }={ \frac{{\sqrt 3 }}{2}\pi {r^3}.}\]

### Example 2.

Find the cylinder with the smallest surface area (Figure \(2a\)).Solution.

The optimal shape of a cylinder at a fixed volume allows to reduce materials cost. Therefore, this problem is important, for example, in the construction of oil storage tanks (Figure \(2a\)).

Let \(H\) be the height of the cylinder and \(R\) be its base radius. The volume and total surface area of the cylinder are calculated by the formulas

\[{V = \pi {R^2}H,}\;\;\;\kern-0.3pt {S = 2\pi {R^2} + 2\pi RH.}\]

We choose the base radius \(R\) as an independent variable. Express \(H\) in terms of \(R\) (at a fixed volume \(V\)):

\[H = \frac{V}{{\pi {R^2}}}.\]

Investigate extreme values of the surface area \(S\left( R \right)\).

\[

{S\left( R \right) = 2\pi {R^2} + 2\pi RH }

= {2\pi {R^2} + 2\pi R \cdot \frac{V}{{\pi {R^2}}} }

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

\]

Calculate the derivative:

\[

{S’\left( R \right) = {\left( {2\pi {R^2} + \frac{{2V}}{R}} \right)^\prime } }

= {4\pi R – \frac{{2V}}{{{R^2}}} }

= {\frac{{4\pi {R^3} – 2V}}{{{R^2}}}.}

\]

Find the stationary points:

\[ {S’\left( R \right) = 0,\;\;}\Rightarrow {\frac{{4\pi {R^3} – 2V}}{{{R^2}}} = 0,\;\;}\Rightarrow {\left\{ {\begin{array}{*{20}{l}} {4\pi {R^3} – 2V = 0}\\ {{R^2} \ne 0} \end{array},} \right.\;\;}\Rightarrow {R = \sqrt[\large 3\normalsize]{{\frac{V}{{2\pi }}}}.} \]

This value of \(R\) corresponds to the minimum surface area \(S\left( R \right),\) as when passing through this point the derivative changes sign from minus to plus.

Calculate the height of the above cylinder:

\[

{H = \frac{V}{{\pi {R^2}}} }

= {\frac{V}{{\pi {{\left( {\sqrt[\large 3\normalsize]{{\frac{V}{{2\pi }}}}} \right)}^2}}} }

= {\frac{{{2^{\large\frac{2}{3}\normalsize}}{\pi ^{\large\frac{2}{3}\normalsize}}V}}{{\pi {V^{\large\frac{2}{3}\normalsize}}}} }

= {\frac{{{2^{\large\frac{2}{3}\normalsize}}{V^{\large\frac{1}{3}\normalsize}}}}{{{\pi ^{\large\frac{1}{3}\normalsize}}}} }

= {\sqrt[\large 3\normalsize]{{\frac{{4V}}{\pi }}}.}

\]

The ratio of the height to the base radius is

\[

{\frac{H}{R} = \frac{{\sqrt[\large 3\normalsize]{{\frac{{4V}}{\pi }}}}}{{\sqrt[\large 3\normalsize]{{\frac{V}{{2\pi }}}}}} }

= {\sqrt[\large 3\normalsize]{{\frac{{4V}}{\pi } \cdot \frac{{2\pi }}{V}}} }

= {\sqrt[\large 3\normalsize]{8} = 2.}

\]

In other words, the height of the cylinder with the smallest surface area should be equal to its diameter, that is an axial section of this cylinder has the form of a square.

### Example 3.

Given a cone with a slant height \(l\) (Figure \(3a\)). Find the largest possible volume of the cone.Solution.

The volume of a cone is

\[V = \frac{1}{3}\pi {R^2}H.\]

By the Pythagorean theorem,

\[{{H^2} + {R^2} = {l^2},}\;\; \Rightarrow {{R^2} = {l^2} – {H^2}.}\]

Hence

\[{V = \frac{\pi }{3}\left( {{l^2} – {H^2}} \right)H }={ \frac{\pi }{3}\left( {{l^2}H – {H^3}} \right) }={ V\left( H \right).}\]

Take the derivative of the function \(V\left( H \right):\)

\[{V^\prime\left( H \right) }={ \left( {\frac{\pi }{3}\left( {{l^2}H – {H^3}} \right)} \right)^\prime }={ \frac{\pi }{3}{l^2} – \pi {H^2}.}\]

Determine the critical value of \(H:\)

\[{V^\prime\left( H \right) = 0,}\;\; \Rightarrow {\frac{\pi }{3}{l^2} – \pi {H^2} = 0,}\;\; \Rightarrow {{H^2} = \frac{{{l^2}}}{3},}\;\; \Rightarrow {H = \frac{l}{{\sqrt 3 }}.}\]

Since

\[V^{\prime\prime}\left( H \right) = – 9H \lt 0,\]

then, by the Second Derivative Test, \(H = \large{\frac{l}{{\sqrt 3 }}}\normalsize\) is a point of local maximum.

The largest volume is equal

\[{{V_{\max }} = \frac{\pi }{3}\left[ {{l^2} \cdot \frac{l}{{\sqrt 3 }} – {{\left( {\frac{l}{{\sqrt 3 }}} \right)}^3}} \right] }={ \frac{\pi }{3}\left( {\frac{{{l^3}}}{{\sqrt 3 }} – \frac{{{l^3}}}{{3\sqrt 3 }}} \right) }={ \frac{\pi }{3} \cdot \frac{{3{l^3} – {l^3}}}{{3\sqrt 3 }} }={ \frac{{2\pi {l^3}}}{{9\sqrt 3 }}.}\]

### Example 4.

Determine the largest volume of a cylinder inscribed in a cone with height \(H\) and base radius \(R\) (Figure \(4a\)).Solution.

We denote the base radius of the inscribed cylinder by \(x,\) and its height by \(y\) (Figure \(4b\)).

From the similarity of triangles \(SKD\) and \(SOB\) we get

\[

{\frac{{KD}}{{OB}} = \frac{{SK}}{{SO}}}\;\;\;\kern-0.3pt

{\text{or}\;\;\;\frac{x}{R} = \frac{{H – y}}{H}.}

\]

The last equation establishes a relationship between the variables \(x\) and \(y.\) Express \(y\) in terms of \(x:\)

\[

{\frac{x}{R} = \frac{{H – y}}{H},\;\;}\Rightarrow

{HX = \left( {H – y} \right)R,\;\;}\Rightarrow

{Hx = HR – Ry,\;\;}\Rightarrow

{y = \frac{{HR – Hx}}{R} = H\left( {1 – \frac{x}{R}} \right).}

\]

The volume of the inscribed cylinder is given by the formula

\[V = \pi {x^2}y.\]

Then

\[

{V\left( x \right) = \pi {x^2}H\left( {1 – \frac{x}{R}} \right) }

= {\pi H\left( {{x^2} – \frac{{{x^3}}}{R}} \right).}

\]

Find the greatest value of the function \(V\left( x \right):\)

\[

{V’\left( x \right) = {\left[ {\pi H\left( {{x^2} – \frac{{{x^3}}}{R}} \right)} \right]^\prime } }

= {\pi H\left( {2x – \frac{{3{x^2}}}{R}} \right);}

\]

\[

{V’\left( x \right) = 0,\;\;}\Rightarrow

{ \pi H\left( {2x – \frac{{3{x^2}}}{R}} \right) = 0,\;\;}\Rightarrow

{ 2x – \frac{{3{x^2}}}{R} = 0,\;\;}\Rightarrow

{ x\left( {2 – \frac{{3x}}{R}} \right) = 0,\;\;}\Rightarrow

{ {x_1} = 0,\;{x_2} = \frac{{2R}}{3}.}

\]

The solution \({x_1} = 0\) corresponds to a zero-volume cylinder and has no physical meaning. When passing through the point \({x_2} = \large\frac{{2R}}{3}\normalsize\) the derivative changes sign from positive to negative. Therefore, \(x = \large\frac{{2R}}{3}\normalsize\) is the point of maximum of the function \(V\left( x \right).\) For this base, the height of the cylinder is given by

\[\require{cancel}

{y = H\left( {1 – \frac{x}{R}} \right) }

= {H\left( {1 – \frac{{2\cancel{R}}}{{3\cancel{R}}}} \right) = \frac{H}{3}.}

\]

Consequently, the largest possible volume of the cylinder inscribed in the given cone is equal to

\[

{{V_{\max }} = \pi {x^2}y }

= {\pi {\left( {\frac{{2R}}{3}} \right)^2} \cdot \frac{H}{3} }

= {\frac{4}{{27}}\pi {R^2}H.}

\]

It is \(\large\frac{4}{9}\normalsize\) of the volume of the cone.

### Example 5.

A coffee filter has a conical shape (Figure \(5a\)). It is made from a circle sector with the central angle \(\alpha\) such that the volume of coffee held in the filter is maximized. Determine the central angle \(\alpha.\)Solution.

The arc length of the sector is given by

\[P = l\alpha ,\]

where \(l\) is the radius of the sector, and the angle \(\alpha\) is measured in radians.

Comparing the two figures we can write the relationship

\[P = 2\pi R,\]

where \(R\) is the base radius of the cone.

It follows from these equations that

\[{P = 2\pi R,}\;\; \Rightarrow {l\alpha = 2\pi R,}\;\; \Rightarrow {R = \frac{{l\alpha }}{{2\pi }}.}\]

The height of the cone is given by

\[{H = \sqrt {{l^2} – {R^2}} }={ \sqrt {{l^2} – {{\left( {\frac{{l\alpha }}{{2\pi }}} \right)}^2}} }={ l\sqrt {1 – \frac{{{\alpha ^2}}}{{4{\pi ^2}}}} }={ \frac{l}{{2\pi }}\sqrt {4{\pi ^2} – {\alpha ^2}} .}\]

Then the volume of the cone is written as

\[{V = \frac{1}{3}\pi {R^2}H }={ \frac{\pi }{3} \cdot {\left( {\frac{{l\alpha }}{{2\pi }}} \right)^2} \cdot \frac{l}{{2\pi }}\sqrt {4{\pi ^2} – {\alpha ^2}} }={ \frac{\pi }{3} \cdot \frac{{{l^2}{\alpha ^2}}}{{4{\pi ^2}}} \cdot \frac{l}{{2\pi }}\sqrt {4{\pi ^2} – {\alpha ^2}} }={ \frac{{{l^3}{\alpha ^2}}}{{24{\pi ^2}}}\sqrt {4{\pi ^2} – {\alpha ^2}} }={ V\left( \alpha \right).}\]

Taking the derivative of the function \(V\left( \alpha \right)\) and equating it to zero, we obtain the equation:

\[\frac{{{l^3}}}{{24{\pi ^2}}} \cdot \frac{{8\alpha {\pi ^2} – 3{\alpha ^2}}}{{\sqrt {4{\pi ^2} – {\alpha ^2}} }} = 0,\]

so the critical points are

\[{8\alpha {\pi ^2} – 3{\alpha ^2} = 0,}\;\; \Rightarrow {\alpha \left( {8{\pi ^2} – 3\alpha } \right) = 0,}\;\; \Rightarrow {\alpha = 0,\,\sqrt {\frac{8}{3}} \pi .}\]

The first solution \(\alpha = 0\) is trivial. The second point \(\alpha = \sqrt {\large{\frac{8}{3}}\normalsize} \pi \) corresponds to the maximum of the function \(V\left( \alpha \right)\) (by the First Derivative Test).

Hence,

\[{\alpha = \sqrt {\frac{8}{3}} \pi }={ 2\pi \sqrt {\frac{2}{3}} }\approx{ 294^\circ.}\]

### Example 6.

Find a cone with the largest volume inscribed in a sphere of radius \(R\) (Figure \(6a\)).Solution.

Consider the axial cross-section of the cone inscribed in the sphere (Figure \(6a\)).

We introduce the following notations: \(H\) is the height of the cone, \(r\) is the base radius of the cone, \(\alpha\) is the angle between the radius of the sphere and the base of the cone. The base radius and height of the cone are connected with the radius of the sphere by the following relationships:

\[{r = R\cos \alpha ,}\;\;\;\kern-0.3pt{H = R\sin \alpha + R.}\]

In such a case, the volume of the cone can be written as

\[

{V = \frac{1}{3}\pi {r^2}H }

= {\frac{1}{3}\pi {\left( {R\cos \alpha } \right)^2}\left( {R\sin \alpha + R} \right) }

= {\frac{1}{3}\pi {R^3}{\cos ^2}\alpha \left( {\sin \alpha + 1} \right).}

\]

where the angle \(\alpha\) varies in the range \(0 < \alpha < {\large\frac{\pi }{2}\normalsize}.\) We differentiate the volume \(V\) with respect to \(\alpha:\)

\[

{V’\left( \alpha \right) }

= {{\left[ {\frac{1}{3}\pi {R^3}{{\cos }^2}\alpha \left( {\sin \alpha + 1} \right)} \right]^\prime } }

= {\frac{1}{3}\pi {R^3}{\left[ {{{\cos }^2}\alpha \left( {\sin \alpha + 1} \right)} \right]^\prime } }

= {{\frac{1}{3}\pi {R^3}\cos \alpha} \cdot\kern0pt {\left[ {{{\cos }^2}\alpha – 2{{\sin }^2}\alpha – 2\sin \alpha } \right]};}

\]

\[

{V’\left( \alpha \right) = 0,\;\;}\Rightarrow

{{\frac{1}{3}\pi {R^3}\cos \alpha}}\cdot\kern0pt{{\left[ {{{\cos }^2}\alpha – 2{{\sin }^2}\alpha – 2\sin \alpha } \right] = 0.}}

\]

\[

{1)\;\cos \alpha = 0,\;\;} \Rightarrow {\alpha = \frac{\pi }{2};}

\]

\[

{2)\;{\cos ^2}\alpha – 2{\sin ^2}\alpha – 2\sin \alpha = 0,\;\;}\Rightarrow

{1 – 3{\sin ^2}\alpha – 2\sin \alpha = 0,\;\;}\Rightarrow

{3{\sin ^2}\alpha + 2\sin \alpha – 1 = 0,\;\;}\Rightarrow

{\sin \alpha = t,\;\;}\Rightarrow

{3{t^2} + 2t – 1 = 0,\;\;}\Rightarrow

{D = 4 – 4 \cdot 3 \cdot \left( { – 1} \right) = 16,\;\;}\Rightarrow

{{t_{1,2}} = \frac{{ – 2 \pm \sqrt {16} }}{6};\;\;}

\]

\[{{t_1} = – 1,\;\;} \Rightarrow {\sin\alpha = – 1,\;\;} \Rightarrow {\alpha = \frac{{3\pi }}{2}},\]

\[{{t_2} = \frac{1}{3},\;\;} \Rightarrow {\sin \alpha = \frac{1}{3}.}\]

As it can be seen, the solution is \(\sin \alpha = \large\frac{1}{3}\normalsize.\) We can check that with increasing \(\alpha\) and passing through this point the derivative changes sign from plus to minus, i.e. here we reach the maximum volume of the cone.

Calculate the cosine of the angle \(\alpha:\)

\[

{\cos\alpha = \sqrt {1 – {{\left( {\frac{1}{3}} \right)}^2}} }

= {\sqrt {1 – \frac{1}{9}} = \frac{{2\sqrt 2 }}{3}.}

\]

Then the base radius and height of the cone of the largest volume have the following values:

\[{r = \frac{{2\sqrt 2 }}{3}R,}\;\;\;\kern-0.3pt{{H = R \cdot \frac{1}{3} + R} = {\frac{4}{3}R.}}\]

The volume of this cone is equal to

\[

{V = \frac{1}{3}\pi {r^2}H }

= {\frac{\pi }{3} \cdot {\left( {\frac{{2\sqrt 2 }}{3}} \right)^2} \cdot \frac{4}{3}R }

= {\frac{\pi }{3} \cdot \frac{8}{9}{R^2} \cdot \frac{4}{3}R }

= {\frac{{32}}{{81}}\pi {R^3},}

\]

which is \(\large\frac{{8}}{{27}}\normalsize\) of the volume of the sphere.

### Example 7.

A piece of cardboard is a rectangle of sides \(a\) and \(b.\) An equal square is cut out from each corner and the sides are folded up to make an open-top rectangular box (Figure \(7a\)). How large the square should be to make the box with the largest possible volume?Solution.

Let \(x\) be the side of the square. The three sides of the rectangular box are equal to \(a – 2x,\) \(b – 2x\) and \(x\). Then the volume of the box is given by

\[{V }={ \left( {a – 2x} \right)\left( {b – 2x} \right)x }={ \left( {ab – 2bx – 2ax + 4{x^2}} \right)x }={ abx – 2{x^2}\left( {a + b} \right) + 4{x^3} }={ V\left( x \right).}\]

Take the derivative of the function \(V\left( x \right):\)

\[{V^\prime\left( x \right) }={ ab – 4\left( {a + b} \right)x + 12{x^2}.}\]

To determine the critical point we need to solve the quadratic equation

\[12{x^2} – 4\left( {a + b} \right)x + ab = 0.\]

Calculate the discriminant:

\[{D }={ 16{\left( {a + b} \right)^2} – 48ab }={ 16\left( {{a^2} + 2ab + {b^2}} \right) – 48ab }={ 16{a^2} + 32ab + 16{b^2} – 48ab }={ 16\left( {{a^2} + ab + {b^2}} \right).}\]

The roots of the equations are

\[{{x_{1,2}} \text{ = }}\kern0pt{ \frac{{4\left( {a + b} \right) \pm \sqrt {16\left( {{a^2} + ab + {b^2}} \right)} }}{{24}} }={ \frac{{a + b \pm \sqrt {{a^2} + ab + {b^2}} }}{6}.}\]

We should take the solution with the minus sign as \(x \lt a\) and \(x \lt b.\) Hence,

\[x = \frac{{a + b – \sqrt {{a^2} + ab + {b^2}} }}{6}.\]