Chapter 21 Maxima and minima of functions of two and three variables-II

Read this section before proceeding to the questions.

21.0.1 Question 5

  1. Find the maximum value of \(f=xyz\) such that \(x+y+z=24\).

Eliminating \(z\) in \(f\), we get

\[\begin{equation*} \begin{split} f &=xy(24-x-y)\\ &= 24xy - x^2y-xy^2\\ f_x &= 24y-2xy-y^2\\ f_y &= 24x-x^2-2xy\\ f_{xx} &= -2y\\ f_{yy} &= -2x\\ f_{xy} &= 24-2x-2y\\ f_{yx} &= 24-2x-2y \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} f_x &= 0\\ 24y-2xy-y^2 &= 0 \tag{21.1} \end{align}\]

\[\begin{align} f_y &= 0\\ 24x-x^2-2xy &= 0 \tag{21.2} \end{align}\]

Solving (21.1) and (21.2), we get

\[\begin{equation*} \begin{split} x &= y\\ x &= 24-y \end{split} \end{equation*}\]

When \(x=y\),

\[\begin{equation*} \begin{split} y &= 0\\ y &= \frac{24}{3}=8 \end{split} \end{equation*}\]

When \(x=24-y\),

\[\begin{equation*} \begin{split} y &= 0\\ y &= 24 \end{split} \end{equation*}\]

Thus there are four extreme points \((0,0),(8,8), (24,0)\) and \((0,24)\).

At \((0,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= 0\\ f_{yy} &= 0\\ f_{xy}=f_{yx} &=24\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} 0 & 24\\ 24 & 0 \end{vmatrix} =-(24)^2 < 0 \end{split} \end{equation*}\]

At \((24,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= 0\\ f_{yy} &= -2\times 24=-48\\ f_{xy}=f_{yx} &=-24\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} 0 & -24\\ -24 & -48 \end{vmatrix} =-(24)^2 < 0 \end{split} \end{equation*}\]

At \((0,24)\),

\[\begin{equation*} \begin{split} f_{xx} &= -2\times 24 =-48 < 0\\ f_{yy} &= 0\\ f_{xy}=f_{yx} &=-24\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} -48 & -24\\ -24 & 0 \end{vmatrix} =-(24)^2 < 0 \end{split} \end{equation*}\]

At points \((0,0), (24,0), (0,24)\), there is neither maximum nor minimum.

At \((8,8)\),

\[\begin{equation*} \begin{split} f_{xx} &= -2\times 8=-16\\ f_{yy} &= -2\times 8=-16\\ f_{xy}=f_{yx} &=-8\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} -16 & -8\\ -8 & -16 \end{vmatrix} = 192 > 0 \end{split} \end{equation*}\]

Thus there is maximum at \((8,8)\). Plugging in these values into \(x+y+z=24\), we get \(z=8\).

Thus at maximum, \(x=y=z=8\) and the value is \(8*8*8=512\).

  1. Find the minimum value of \(u=x^2 + xy + y^2 + 3z^2\) such that \(x+2y+4z=60\).

Here \(u=x^2 + xy + y^2 + 3z^2\) and \(\phi=x+2y+4z-60\).

Let \(F=u+\lambda \phi=x^2 + xy + y^2 + 3z^2 +\lambda (x+2y+4z-60)\). So

\[\begin{equation*} \begin{split} F_x &= 2x+y+\lambda\\ F_y &= x+2y+2\lambda\\ F_z &= 6z + 4\lambda \\ F_{\lambda} &= x+2y+4z-60 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} F_x &= 0\\ 2x+y+\lambda &= 0 \tag{21.3} \end{align}\]

\[\begin{align} F_y &= 0\\ x+2y+2\lambda &= 0 \tag{21.4} \end{align}\]

\[\begin{align} F_z &= 0\\ 6z + 4\lambda &= 0 \tag{21.5} \end{align}\]

\[\begin{align} F_{\lambda} &= 0\\ x+2y+4z-60 &= 0 \tag{21.6} \end{align}\]

From (21.3), (21.4), (21.5) and (21.6), we get \(x=0, y=\frac{90}{7}, z=\frac{60}{7}\).

Taking partial derivatives,

\[\begin{equation*} \begin{split} u_x &= 2x+y\\ u_y &= x+2y\\ u_z &= 6z\\ u_{xx} &= 2\\ u_{xy} &= 1\\ u_{xz} &= 0\\ u_{yx} &= 1\\ u_{yy} &= 2\\ u_{yz} &= 0 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} u_{zx} &= 0\\ u_{zy} &= 0\\ u_{zz} &= 6\\ \phi_x &= 1\\ \phi_y &= 2\\ \phi_z &= 4 \end{split} \end{equation*}\]

At \(\left(0,\frac{90}{7}, \frac{60}{7}\right)\), calculating bordered Hessian determinants,

\[\begin{equation*} \begin{split} \begin{vmatrix}\widetilde{H1}\end{vmatrix}= \begin{vmatrix} 0 & \phi_{x} \\ \phi_{x} & u_{xx} \end{vmatrix} = \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} =-1 < 0 \end{split} \end{equation*}\]

\[ \begin{vmatrix}\widetilde{H2}\end{vmatrix}= \begin{vmatrix} 0 & \phi_{x} & \phi_{y} \\ \phi_{x} & u_{xx} & u_{xy} \\ \phi_{y} & u_{yx} & u_{yy} \end{vmatrix} = \begin{vmatrix} 0 & 1 & 2 \\ 1 & 2 & 1\\ 2 & 1 & 2 \end{vmatrix} =-6 < 0 \]

\[ \begin{vmatrix}\widetilde{H3}\end{vmatrix}= \begin{vmatrix} 0 & \phi_{x} & \phi_{y} & \phi_{z} \\ \phi_{x} & u_{xx} & u_{xy} & u_{xz} \\ \phi_{y} & u_{yx} & u_{yy} & u_{yz} \\ \phi_{z} & u_{zx} & u_{zy} & u_{zz} \end{vmatrix} = \begin{vmatrix} 0 & 1 & 2 & 4 \\ 1 & 2 & 1 & 0 \\ 2 & 1 & 2 & 0\\ 4 & 0 & 0 & 6 \end{vmatrix} = -84 < 0 \]

The way I calculated determinant in 4x4 matrix is using R code:

x=c(0,1,2,4,1,2,1,0,2,1,2,0,4,0,0,6)
y= matrix(x, nrow = 4, ncol = 4, byrow = TRUE)
det(y)

Thus all bordered Hessian determinants \(\begin{vmatrix}\widetilde{H1}\end{vmatrix}, \begin{vmatrix}\widetilde{H2}\end{vmatrix}\) and \(\begin{vmatrix}\widetilde{H3}\end{vmatrix}\) are \(<0\). Hence \(f\) has minimum at \(\left(0,\frac{90}{7}, \frac{60}{7}\right)\) and the value is

\[\begin{equation*} \begin{split} u &= \left(\frac{90}{7}\right)^2 + 3\times \left(\frac{60}{7}\right)^2\\ &= \frac{2700}{7} \end{split} \end{equation*}\]

The way I add these fractions is using Raku code:

say (8100/49+10800/49).raku
  1. Find the extreme value of \(\phi = x^2 + y^2 + z^2\) such that \(x+z=1\) and \(2y+z=2\).

Given, \(\phi=x^2+y^2+z^2\). Let \(\beta=x+z-1, \psi=2y+z-2\), then

\[\begin{equation*} \begin{split} F &=\phi+ \lambda_1 \beta +\lambda_2\psi\\ &= x^2+y^2+z^2 + \lambda_1(x+z-1)+\lambda_2(2y+z-2)\\ F_x &= 2x + \lambda_1 \\ F_y &= 2y+2\lambda_2 \\ F_z &= 2z + \lambda_1 + \lambda_2\\ F_{\lambda_1} &= x+z-1\\ F_{\lambda_2} &= 2y + z-2 \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} F_x &= 0\\ 2x + \lambda_1 &= 0 \tag{21.7} \end{align}\]

\[\begin{align} F_y &= 0\\ 2y + 2\lambda_2 &= 0 \tag{21.8} \end{align}\]

\[\begin{align} F_z &= 0\\ 2z + \lambda_1 + \lambda_2 &= 0 \tag{21.9} \end{align}\]

\[\begin{align} F_{\lambda_1} &= 0\\ x + z-1 &= 0 \tag{21.10} \end{align}\]

\[\begin{align} F_{\lambda_2} &= 0\\ 2y + z -2 &= 0 \tag{21.11} \end{align}\]

Multiplying (21.7) by \(2\) and (21.9) by \(2\), we get

\[\begin{align} 4x + 2\lambda_1 &= 0 \tag{21.12} \end{align}\]

\[\begin{align} 4z+ 2\lambda_1+ 2\lambda_2 &= 0 \tag{21.13} \end{align}\]

Adding (21.12) and (21.8) and subtracting (21.13), we have

\[\begin{align} 4x+2y-4z &= 0 \tag{21.14} \end{align}\]

Solving (21.10), (21.11) and (21.14), \(x=\frac{1}{3}, y=\frac{2}{3}\) and \(z=\frac{2}{3}\).

The extreme value is thus

\[\begin{equation*} \begin{split} \phi\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right) &= \left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2\\ &= \frac{9}{9}\\ &= 1 \end{split} \end{equation*}\]

Lagrange’s method of undetermined multipliers can help only in determining the extreme point but it cannot say definitely whether it has a maximum or a minimum value at that point.

21.0.2 Question 6

Before solving this question read Stewart’s Calculus Early Transcendentals pp 1020.

Find the minimum value of \(x^2 + y^2 + z^2\) when subjected to the condition \(x+y+z-1=0\) and \(xyz+1=0\).

Let \(f=x^2 + y^2 + z^2, g=x+y+z-1\) and \(h=xyz+1\). We are trying to minimise the function \(f\) subject to the constraints \(g\) and \(h\). Calculating the gradients of each

\[\begin{equation*} \begin{split} \nabla f &= (2x,2y,2z)\\ \nabla g &= (1,1,1)\\ \nabla h &= (yz, xz, xy) \end{split} \end{equation*}\]

The Lagrange condition is \(\nabla f= \lambda \nabla g + \mu \nabla h\), so we solve the equations

\[\begin{align} 2x &= \lambda + \mu yz \tag{21.15} \end{align}\]

\[\begin{align} 2y &= \lambda + \mu xz \tag{21.16} \end{align}\]

\[\begin{align} 2z &= \lambda + \mu xy \tag{21.17} \end{align}\]

\[\begin{align} x+y+z &= 1 \tag{21.18} \end{align}\]

\[\begin{align} xyz + 1 &= 0 \tag{21.19} \end{align}\]

Subtracting (21.16) from (21.15),

\[\begin{equation*} \begin{split} 2x-2y &= \mu z (y-x)\\ \mu z (y-x) + 2(y-x) &= 0\\ (y-x)(\mu z + 2) &= 0\\ y = x, & \qquad z=-\frac{2}{\mu} \end{split} \end{equation*}\]

Putting these values in (21.18),

\[\begin{align} 2x-\frac{2}{\mu} &= 1 \tag{21.20} \end{align}\]

and (21.19),

\[\begin{align} -\frac{2x^2}{\mu} + 1 &= 0 \tag{21.21} \end{align}\]

Solving (21.20) and (21.21),

\[\begin{equation*} \begin{split} x^2(1-2x) + 1 &= 0\\ 2x^3-x^2-1 &= 0\\ x &= 1, -\frac{1}{4}-\frac{\sqrt{7}i}{4}, -\frac{1}{4}+\frac{\sqrt{7}i}{4} \end{split} \end{equation*}\]

Taking \(x=1\), \(y=1, z=-1\). The extreme point is thus \((1,1,-1)\).

Subtracting (21.17) from (21.15),

\[\begin{equation*} \begin{split} 2x-2z &= \mu y (z-x)\\ (x-z)(2+ \mu y) &= 0\\ x = z, & \qquad y=-\frac{2}{\mu} \end{split} \end{equation*}\]

Putting these values in (21.18),

\[\begin{align} 2z-\frac{2}{\mu} &= 1 \tag{21.22} \end{align}\]

and (21.19),

\[\begin{align} -\frac{2z^2}{\mu} + 1 &= 0 \tag{21.23} \end{align}\]

Solving (21.22) and (21.23),

\[\begin{equation*} \begin{split} z^2(1-2z) + 1 &= 0\\ 2z^3-z^2-1 &= 0\\ z &= 1, -\frac{1}{4}-\frac{\sqrt{7}i}{4}, -\frac{1}{4}+\frac{\sqrt{7}i}{4} \end{split} \end{equation*}\]

Taking \(z=1\), \(x=1,y=-1\). The extreme point is thus \((1,-1,1)\).

Subtracting (21.17) from (21.16),

\[\begin{equation*} \begin{split} 2y-2z &= \mu x (z-y)\\ \mu x (z-y) + 2(z-y) &= 0\\ (z-y)(\mu x + 2) &= 0\\ z = y, & \qquad x=-\frac{2}{\mu} \end{split} \end{equation*}\]

Putting these values in (21.18),

\[\begin{align} 2y-\frac{2}{\mu} &= 1 \tag{21.24} \end{align}\]

and (21.19),

\[\begin{align} -\frac{2y^2}{\mu} + 1 &= 0 \tag{21.25} \end{align}\]

Solving (21.24) and (21.25),

\[\begin{equation*} \begin{split} y^2(1-2y) + 1 &= 0\\ 2y^3-y^2-1 &= 0\\ y &= 1, -\frac{1}{4}-\frac{\sqrt{7}i}{4}, -\frac{1}{4}+\frac{\sqrt{7}i}{4} \end{split} \end{equation*}\]

Taking \(y=1\), \(x=-1, z=1\). The extreme point is thus \((-1,1,1)\).

Thus there are three extreme points:

  1. \((1,1,-1)\)
  2. \((1,-1,1)\)
  3. \((-1,1,1)\)

The function has the value \(3\) at each of these points. The extreme value is thus \(3\).

21.0.3 Question 7

If the sum of three positive numbers is the cube of \(2\), what is the maximum value of their product.

Let \(x,y,z\) be three positive numbers. And we are given \(x+y+z=8\). We have to maximise the product \(f=xyz\). Eliminating \(z\) in \(f\), we get

\[\begin{equation*} \begin{split} f &=xy(8-x-y)\\ &= 8xy - x^2y-xy^2\\ f_x &= 8y-2xy-y^2\\ f_y &= 8x-x^2-2xy\\ f_{xx} &= -2y\\ f_{yy} &= -2x\\ f_{xy} &= 8-2x-2y\\ f_{yx} &= 8-2x-2y \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} f_x &= 0\\ 8y-2xy-y^2 &= 0 \tag{21.26} \end{align}\]

\[\begin{align} f_y &= 0\\ 8x-x^2-2xy &= 0 \tag{21.27} \end{align}\]

Solving (21.26) and (21.27), we get

\[\begin{equation*} \begin{split} x &= y\\ x &= 8-y \end{split} \end{equation*}\]

When \(x=y\),

\[\begin{equation*} \begin{split} y &= 0\\ y &= \frac{8}{3} \end{split} \end{equation*}\]

When \(x=8-y\),

\[\begin{equation*} \begin{split} y &= 0\\ y &= 8 \end{split} \end{equation*}\]

Thus there are four extreme points \((0,0),\left(\frac{8}{3}, \frac{8}{3}\right), (8,0)\) and \((0,8)\).

At \((0,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= 0\\ f_{yy} &= 0\\ f_{xy}=f_{yx} &=8\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} 0 & 8\\ 8 & 0 \end{vmatrix} =-8^2 < 0 \end{split} \end{equation*}\]

At \((8,0)\),

\[\begin{equation*} \begin{split} f_{xx} &= 0\\ f_{yy} &= -2\times 8=-16\\ f_{xy}=f_{yx} &=-8\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} 0 & -8\\ -8 & -16 \end{vmatrix} =-8^2 < 0 \end{split} \end{equation*}\]

At \((0,8)\),

\[\begin{equation*} \begin{split} f_{xx} &= -2\times 8=-16 < 0\\ f_{yy} &= 0\\ f_{xy}=f_{yx} &=-8\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} -16 & -8\\ -8 & 0 \end{vmatrix} =-8^2 < 0 \end{split} \end{equation*}\]

At points \((0,0), (8,0), (0,8)\), there is neither maximum nor minimum.

At \(\left(\frac{8}{3}, \frac{8}{3}\right)\),

\[\begin{equation*} \begin{split} f_{xx} &= \frac{-16}{3}\\ f_{yy} &= \frac{-16}{3}\\ f_{xy}=f_{yx} &=\frac{-8}{3}\\ \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix} =\begin{vmatrix} \frac{-16}{3} & \frac{-8}{3}\\ \frac{-8}{3} & \frac{-16}{3} \end{vmatrix} = \frac{8^2}{3} > 0 \end{split} \end{equation*}\]

Thus there is maximum at \(\left(\frac{8}{3}, \frac{8}{3}\right)\). Plugging in these values into \(x+y+z=8\), we get \(z=\frac{8}{3}\).

Thus at maximum, \(x=y=z=\frac{8}{3}\) and maximum value is \(\frac{8}{3}\times\frac{8}{3}\times\frac{8}{3}=\frac{512}{27}\).

21.0.4 Question 8

If the sum of the dimensions of a rectangular swimming pool is given, prove that the amount of water in the pool is maximum when it is a cube.

Let the dimensions of the swimming pool be \(x\), \(y\) and \(z\). Then the amount of water in the pool will be \(V=xyz\) where \(V=\) volume of water. But what we are given is the sum of the dimensions of the pool i.e. \(x+y+z=a\). Within this constraint we have to maximise the amount of water i.e. volume \(V\).

In short, maximise \(V=xyz\) given \(x+y+z=a\). Eliminating \(z\) in \(V\), we get

\[\begin{equation*} \begin{split} V &=xy(a-x-y)\\ &= axy - x^2y-xy^2\\ V_x &= ay-2xy-y^2\\ V_y &= ax-x^2-2xy\\ V_{xx} &= -2y\\ V_{yy} &= -2x\\ V_{xy} &= a-2x-2y\\ V_{yx} &= a-2x-2y \end{split} \end{equation*}\]

For extreme values,

\[\begin{align} V_x &= 0\\ ay-2xy-y^2 &= 0 \tag{21.28} \end{align}\]

\[\begin{align} V_y &= 0\\ ax-x^2-2xy &= 0 \tag{21.29} \end{align}\]

Solving (21.28) and (21.29), we get

\[\begin{equation*} \begin{split} x &= y\\ x &= a-y \end{split} \end{equation*}\]

When \(x=y\),

\[\begin{equation*} \begin{split} y &= 0\\ y &= \frac{a}{3} \end{split} \end{equation*}\]

When \(x=a-y\),

\[\begin{equation*} \begin{split} y &= 0\\ y &= a \end{split} \end{equation*}\]

Thus there are four extreme points \((0,0),\left(\frac{a}{3}, \frac{a}{3}\right), (a,0)\) and \((0,a)\).

At \((0,0)\),

\[\begin{equation*} \begin{split} V_{xx} &= 0\\ V_{yy} &= 0\\ V_{xy}=V_{yx} &=a\\ \begin{vmatrix} V_{xx} & V_{xy} \\ V_{yx} & V_{yy} \end{vmatrix} =\begin{vmatrix} 0 & a\\ a & 0 \end{vmatrix} =-a^2 < 0 \end{split} \end{equation*}\]

At \((a,0)\),

\[\begin{equation*} \begin{split} V_{xx} &= 0\\ V_{yy} &= -2a\\ V_{xy}=V_{yx} &=-a\\ \begin{vmatrix} V_{xx} & V_{xy} \\ V_{yx} & V_{yy} \end{vmatrix} =\begin{vmatrix} 0 & -a\\ -a & -2a \end{vmatrix} =-a^2 < 0 \end{split} \end{equation*}\]

At \((0,a)\),

\[\begin{equation*} \begin{split} V_{xx} &= -2a < 0\\ V_{yy} &= 0\\ V_{xy}=V_{yx} &=-a\\ \begin{vmatrix} V_{xx} & V_{xy} \\ V_{yx} & V_{yy} \end{vmatrix} =\begin{vmatrix} -2a & -a\\ -a & 0 \end{vmatrix} =-a^2 < 0 \end{split} \end{equation*}\]

At points \((0,0), (a,0), (0,a)\), there is neither maximum nor minimum.

At \(\left(\frac{a}{3}, \frac{a}{3}\right)\),

\[\begin{equation*} \begin{split} V_{xx} &= \frac{-2a}{3}\\ V_{yy} &= \frac{-2a}{3}\\ V_{xy}=V_{yx} &=\frac{-a}{3}\\ \begin{vmatrix} V_{xx} & V_{xy} \\ V_{yx} & V_{yy} \end{vmatrix} =\begin{vmatrix} \frac{-2a}{3} & \frac{-a}{3}\\ \frac{-a}{3} & \frac{-2a}{3} \end{vmatrix} = \frac{a^2}{3} > 0 \end{split} \end{equation*}\]

Thus there is maximum at \(\left(\frac{a}{3}, \frac{a}{3}\right)\). Plugging in these values into \(x+y+z=a\), we get \(z=\frac{a}{3}\).

Thus at maximum, \(x=y=z=\frac{a}{3}\) which is indeed a cube.

21.0.5 Question 9

Find a point within a triangle such that the sum of the squares of its distances from the three angular points is minimum.

Let the vertices of the triangle be \((x_1,y_1), (x_2,y_2)\) and \((x_3,y_3)\) and the point inside triangle be \((x,y)\).

Let \(d\) be the sum of the squares of the distances of \((x,y)\) from three angular points (vertices).

\[\begin{equation*} \begin{split} d &= [(x-x_1)^2 + (y-y_1)^2] + [(x-x_2)^2 + (y-y_2)^2] + [(x-x_3)^2 + (y-y_3)^2]\\ \end{split} \end{equation*}\]

What we have to do is minimize the \(d\).

\[\begin{equation*} \begin{split} d_x &= 2(x-x_1) + 2(x-x_2) + 2(x-x_3)\\ d_y &= 2(y-y_1) + 2(y-y_2) + 2(y-y_3) \end{split} \end{equation*}\]

For extreme values,

\[\begin{equation*} \begin{split} d_x &= 0\\ 2(x-x_1) + 2(x-x_2) + 2(x-x_3) &= 0\\ x &= \frac{x_1 + x_2 + x_3}{3} \end{split} \end{equation*}\]

and

\[\begin{equation*} \begin{split} d_y &= 0\\ 2(y-y_1) + 2(y-y_2) + 2(y-y_3) &= 0\\ y &= \frac{y_1 + y_2 + y_3}{3} \end{split} \end{equation*}\]

Thus \(d\) has extreme value at \(\left(\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3}\right)\).

Now calculating partial second derivatives,

\[\begin{equation*} \begin{split} d_{xx} &= 6\\ d_{xy} &= 0\\ d_{yy} &= 6\\ d_{yx} &= 0 \end{split} \end{equation*}\]

Thus \(d_{xx}=6 > 0\) and

\[ \begin{vmatrix} d_{xx} & d_{xy} \\ d_{yx} & d_{yy} \end{vmatrix} = \begin{vmatrix} 6 & 0\\ 0 & 6 \end{vmatrix} =36 >0 \]

Thus \(d\) is minimum at \(\left(\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3}\right)\). Hence, a point within a triangle such that the sum of the squares of its distances from the three angular points is minimum is \(\left(\frac{x_1 + x_2 + x_3}{3},\frac{y_1 + y_2 + y_3}{3}\right)\).