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

Find the maximum value of $f = x y z$ such that $x + y + z = 24$ .

Eliminating $z$ in $f$ , we get

$\begin{aligned} f & = x y (24 - x - y) \\ = 24 x y - x^{2} y - x y^{2} \\ f_{x} & = 24 y - 2 x y - y^{2} \\ f_{y} & = 24 x - x^{2} - 2 x y \\ f_{x x} & = - 2 y \\ f_{y y} & = - 2 x \\ f_{x y} & = 24 - 2 x - 2 y \\ f_{y x} & = 24 - 2 x - 2 y \end{aligned}$

For extreme values,

$\begin{aligned} f_{x} & = 0 \\ (21.1) & 24 y - 2 x y - y^{2} & = 0 \end{aligned}$

$\begin{aligned} f_{y} & = 0 \\ (21.2) & 24 x - x^{2} - 2 x y & = 0 \end{aligned}$

Solving (21.1) and (21.2), we get

$\begin{aligned} x & = y \\ x & = 24 - y \end{aligned}$

When $x = y$ ,

$\begin{aligned} y & = 0 \\ y & = \frac{24}{3} = 8 \end{aligned}$

When $x = 24 - y$ ,

$\begin{aligned} y & = 0 \\ y & = 24 \end{aligned}$

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

At $(0, 0)$ ,

$\begin{aligned} f_{x x} & = 0 \\ f_{y y} & = 0 \\ f_{x y} = f_{y x} & = 24 \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} 0 & 24 \\ 24 & 0 \end{matrix} | = - (24)^{2} < 0 \end{aligned}$

At $(24, 0)$ ,

$\begin{aligned} f_{x x} & = 0 \\ f_{y y} & = - 2 \times 24 = - 48 \\ f_{x y} = f_{y x} & = - 24 \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} 0 & - 24 \\ - 24 & - 48 \end{matrix} | = - (24)^{2} < 0 \end{aligned}$

At $(0, 24)$ ,

$\begin{aligned} f_{x x} & = - 2 \times 24 = - 48 < 0 \\ f_{y y} & = 0 \\ f_{x y} = f_{y x} & = - 24 \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} - 48 & - 24 \\ - 24 & 0 \end{matrix} | = - (24)^{2} < 0 \end{aligned}$

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

At $(8, 8)$ ,

$\begin{aligned} f_{x x} & = - 2 \times 8 = - 16 \\ f_{y y} & = - 2 \times 8 = - 16 \\ f_{x y} = f_{y x} & = - 8 \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} - 16 & - 8 \\ - 8 & - 16 \end{matrix} | = 192 > 0 \end{aligned}$

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 $8 * 8 * 8 = 512$ .

Find the minimum value of $u = x^{2} + x y + y^{2} + 3 z^{2}$ such that $x + 2 y + 4 z = 60$ .

Here $u = x^{2} + x y + y^{2} + 3 z^{2}$ and $ϕ = x + 2 y + 4 z - 60$ .

Let $F = u + λ ϕ = x^{2} + x y + y^{2} + 3 z^{2} + λ (x + 2 y + 4 z - 60)$ . So

$\begin{aligned} F_{x} & = 2 x + y + λ \\ F_{y} & = x + 2 y + 2 λ \\ F_{z} & = 6 z + 4 λ \\ F_{λ} & = x + 2 y + 4 z - 60 \end{aligned}$

For extreme values,

$\begin{aligned} F_{x} & = 0 \\ (21.3) & 2 x + y + λ & = 0 \end{aligned}$

$\begin{aligned} F_{y} & = 0 \\ (21.4) & x + 2 y + 2 λ & = 0 \end{aligned}$

$\begin{aligned} F_{z} & = 0 \\ (21.5) & 6 z + 4 λ & = 0 \end{aligned}$

$\begin{aligned} F_{λ} & = 0 \\ (21.6) & x + 2 y + 4 z - 60 & = 0 \end{aligned}$

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,

$\begin{aligned} u_{x} & = 2 x + y \\ u_{y} & = x + 2 y \\ u_{z} & = 6 z \\ u_{x x} & = 2 \\ u_{x y} & = 1 \\ u_{x z} & = 0 \\ u_{y x} & = 1 \\ u_{y y} & = 2 \\ u_{y z} & = 0 \end{aligned}$

$\begin{aligned} u_{z x} & = 0 \\ u_{z y} & = 0 \\ u_{z z} & = 6 \\ ϕ_{x} & = 1 \\ ϕ_{y} & = 2 \\ ϕ_{z} & = 4 \end{aligned}$

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

$\begin{aligned} | \begin{matrix} \tilde{H 1} \end{matrix} | = | \begin{matrix} 0 & ϕ_{x} \\ ϕ_{x} & u_{x x} \end{matrix} | = | \begin{matrix} 0 & 1 \\ 1 & 2 \end{matrix} | = - 1 < 0 \end{aligned}$

$| \begin{matrix} \tilde{H 2} \end{matrix} | = | \begin{matrix} 0 & ϕ_{x} & ϕ_{y} \\ ϕ_{x} & u_{x x} & u_{x y} \\ ϕ_{y} & u_{y x} & u_{y y} \end{matrix} | = | \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 2 \end{matrix} | = - 6 < 0$

$| \begin{matrix} \tilde{H 3} \end{matrix} | = | \begin{matrix} 0 & ϕ_{x} & ϕ_{y} & ϕ_{z} \\ ϕ_{x} & u_{x x} & u_{x y} & u_{x z} \\ ϕ_{y} & u_{y x} & u_{y y} & u_{y z} \\ ϕ_{z} & u_{z x} & u_{z y} & u_{z z} \end{matrix} | = | \begin{matrix} 0 & 1 & 2 & 4 \\ 1 & 2 & 1 & 0 \\ 2 & 1 & 2 & 0 \\ 4 & 0 & 0 & 6 \end{matrix} | = - 84 < 0$

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 $| \begin{matrix} \tilde{H 1} \end{matrix} |, | \begin{matrix} \tilde{H 2} \end{matrix} |$ and $| \begin{matrix} \tilde{H 3} \end{matrix} |$ are $< 0$ . Hence $f$ has minimum at $(0, \frac{90}{7}, \frac{60}{7})$ and the value is

$\begin{aligned} u & = {(\frac{90}{7})}^{2} + 3 \times {(\frac{60}{7})}^{2} \\ = \frac{2700}{7} \end{aligned}$

The way I add these fractions is using Raku code:

say (8100/49+10800/49).raku

Find the extreme value of $ϕ = x^{2} + y^{2} + z^{2}$ such that $x + z = 1$ and $2 y + z = 2$ .

Given, $ϕ = x^{2} + y^{2} + z^{2}$ . Let $β = x + z - 1, ψ = 2 y + z - 2$ , then

$\begin{aligned} F & = ϕ + λ_{1} β + λ_{2} ψ \\ = x^{2} + y^{2} + z^{2} + λ_{1} (x + z - 1) + λ_{2} (2 y + z - 2) \\ F_{x} & = 2 x + λ_{1} \\ F_{y} & = 2 y + 2 λ_{2} \\ F_{z} & = 2 z + λ_{1} + λ_{2} \\ F_{λ_{1}} & = x + z - 1 \\ F_{λ_{2}} & = 2 y + z - 2 \end{aligned}$

For extreme values,

$\begin{aligned} F_{x} & = 0 \\ (21.7) & 2 x + λ_{1} & = 0 \end{aligned}$

$\begin{aligned} F_{y} & = 0 \\ (21.8) & 2 y + 2 λ_{2} & = 0 \end{aligned}$

$\begin{aligned} F_{z} & = 0 \\ (21.9) & 2 z + λ_{1} + λ_{2} & = 0 \end{aligned}$

$\begin{aligned} F_{λ_{1}} & = 0 \\ (21.10) & x + z - 1 & = 0 \end{aligned}$

$\begin{aligned} F_{λ_{2}} & = 0 \\ (21.11) & 2 y + z - 2 & = 0 \end{aligned}$

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

$\begin{aligned} (21.12) & 4 x + 2 λ_{1} & = 0 \end{aligned}$

$\begin{aligned} (21.13) & 4 z + 2 λ_{1} + 2 λ_{2} & = 0 \end{aligned}$

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

$\begin{aligned} (21.14) & 4 x + 2 y - 4 z & = 0 \end{aligned}$

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

$\begin{aligned} ϕ (\frac{1}{3}, \frac{2}{3}, \frac{2}{3}) & = {(\frac{1}{3})}^{2} + {(\frac{2}{3})}^{2} + {(\frac{2}{3})}^{2} \\ = \frac{9}{9} \\ = 1 \end{aligned}$

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 $x y z + 1 = 0$ .

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

$\begin{aligned} \nabla f & = (2 x, 2 y, 2 z) \\ \nabla g & = (1, 1, 1) \\ \nabla h & = (y z, x z, x y) \end{aligned}$

The Lagrange condition is $\nabla f = λ \nabla g + μ \nabla h$ , so we solve the equations

$\begin{aligned} (21.15) & 2 x & = λ + μ y z \end{aligned}$

$\begin{aligned} (21.16) & 2 y & = λ + μ x z \end{aligned}$

$\begin{aligned} (21.17) & 2 z & = λ + μ x y \end{aligned}$

$\begin{aligned} (21.18) & x + y + z & = 1 \end{aligned}$

$\begin{aligned} (21.19) & x y z + 1 & = 0 \end{aligned}$

Subtracting (21.16) from (21.15),

$\begin{aligned} 2 x - 2 y & = μ z (y - x) \\ μ z (y - x) + 2 (y - x) & = 0 \\ (y - x) (μ z + 2) & = 0 \\ y = x, & z = - \frac{2}{μ} \end{aligned}$

Putting these values in (21.18),

$\begin{aligned} (21.20) & 2 x - \frac{2}{μ} & = 1 \end{aligned}$

and (21.19),

$\begin{aligned} (21.21) & - \frac{2 x^{2}}{μ} + 1 & = 0 \end{aligned}$

Solving (21.20) and (21.21),

$\begin{aligned} x^{2} (1 - 2 x) + 1 & = 0 \\ 2 x^{3} - x^{2} - 1 & = 0 \\ x & = 1, - \frac{1}{4} - \frac{\sqrt{7} i}{4}, - \frac{1}{4} + \frac{\sqrt{7} i}{4} \end{aligned}$

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

Subtracting (21.17) from (21.15),

$\begin{aligned} 2 x - 2 z & = μ y (z - x) \\ (x - z) (2 + μ y) & = 0 \\ x = z, & y = - \frac{2}{μ} \end{aligned}$

Putting these values in (21.18),

$\begin{aligned} (21.22) & 2 z - \frac{2}{μ} & = 1 \end{aligned}$

and (21.19),

$\begin{aligned} (21.23) & - \frac{2 z^{2}}{μ} + 1 & = 0 \end{aligned}$

Solving (21.22) and (21.23),

$\begin{aligned} z^{2} (1 - 2 z) + 1 & = 0 \\ 2 z^{3} - z^{2} - 1 & = 0 \\ z & = 1, - \frac{1}{4} - \frac{\sqrt{7} i}{4}, - \frac{1}{4} + \frac{\sqrt{7} i}{4} \end{aligned}$

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

Subtracting (21.17) from (21.16),

$\begin{aligned} 2 y - 2 z & = μ x (z - y) \\ μ x (z - y) + 2 (z - y) & = 0 \\ (z - y) (μ x + 2) & = 0 \\ z = y, & x = - \frac{2}{μ} \end{aligned}$

Putting these values in (21.18),

$\begin{aligned} (21.24) & 2 y - \frac{2}{μ} & = 1 \end{aligned}$

and (21.19),

$\begin{aligned} (21.25) & - \frac{2 y^{2}}{μ} + 1 & = 0 \end{aligned}$

Solving (21.24) and (21.25),

$\begin{aligned} y^{2} (1 - 2 y) + 1 & = 0 \\ 2 y^{3} - y^{2} - 1 & = 0 \\ y & = 1, - \frac{1}{4} - \frac{\sqrt{7} i}{4}, - \frac{1}{4} + \frac{\sqrt{7} i}{4} \end{aligned}$

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

Thus there are three extreme points:

$(1, 1, - 1)$
$(1, - 1, 1)$
$(- 1, 1, 1)$

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 $f = x y z$ . Eliminating $z$ in $f$ , we get

$\begin{aligned} f & = x y (8 - x - y) \\ = 8 x y - x^{2} y - x y^{2} \\ f_{x} & = 8 y - 2 x y - y^{2} \\ f_{y} & = 8 x - x^{2} - 2 x y \\ f_{x x} & = - 2 y \\ f_{y y} & = - 2 x \\ f_{x y} & = 8 - 2 x - 2 y \\ f_{y x} & = 8 - 2 x - 2 y \end{aligned}$

For extreme values,

$\begin{aligned} f_{x} & = 0 \\ (21.26) & 8 y - 2 x y - y^{2} & = 0 \end{aligned}$

$\begin{aligned} f_{y} & = 0 \\ (21.27) & 8 x - x^{2} - 2 x y & = 0 \end{aligned}$

Solving (21.26) and (21.27), we get

$\begin{aligned} x & = y \\ x & = 8 - y \end{aligned}$

When $x = y$ ,

$\begin{aligned} y & = 0 \\ y & = \frac{8}{3} \end{aligned}$

When $x = 8 - y$ ,

$\begin{aligned} y & = 0 \\ y & = 8 \end{aligned}$

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

At $(0, 0)$ ,

$\begin{aligned} f_{x x} & = 0 \\ f_{y y} & = 0 \\ f_{x y} = f_{y x} & = 8 \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} 0 & 8 \\ 8 & 0 \end{matrix} | = - 8^{2} < 0 \end{aligned}$

At $(8, 0)$ ,

$\begin{aligned} f_{x x} & = 0 \\ f_{y y} & = - 2 \times 8 = - 16 \\ f_{x y} = f_{y x} & = - 8 \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} 0 & - 8 \\ - 8 & - 16 \end{matrix} | = - 8^{2} < 0 \end{aligned}$

At $(0, 8)$ ,

$\begin{aligned} f_{x x} & = - 2 \times 8 = - 16 < 0 \\ f_{y y} & = 0 \\ f_{x y} = f_{y x} & = - 8 \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} - 16 & - 8 \\ - 8 & 0 \end{matrix} | = - 8^{2} < 0 \end{aligned}$

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

At $(\frac{8}{3}, \frac{8}{3})$ ,

$\begin{aligned} f_{x x} & = \frac{- 16}{3} \\ f_{y y} & = \frac{- 16}{3} \\ f_{x y} = f_{y x} & = \frac{- 8}{3} \\ | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = | \begin{matrix} \frac{- 16}{3} & \frac{- 8}{3} \\ \frac{- 8}{3} & \frac{- 16}{3} \end{matrix} | = \frac{8^{2}}{3} > 0 \end{aligned}$

Thus there is maximum at $(\frac{8}{3}, \frac{8}{3})$ . 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 $V = x y z$ 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 $V = x y z$ given $x + y + z = a$ . Eliminating $z$ in $V$ , we get

$\begin{aligned} V & = x y (a - x - y) \\ = a x y - x^{2} y - x y^{2} \\ V_{x} & = a y - 2 x y - y^{2} \\ V_{y} & = a x - x^{2} - 2 x y \\ V_{x x} & = - 2 y \\ V_{y y} & = - 2 x \\ V_{x y} & = a - 2 x - 2 y \\ V_{y x} & = a - 2 x - 2 y \end{aligned}$

For extreme values,

$\begin{aligned} V_{x} & = 0 \\ (21.28) & a y - 2 x y - y^{2} & = 0 \end{aligned}$

$\begin{aligned} V_{y} & = 0 \\ (21.29) & a x - x^{2} - 2 x y & = 0 \end{aligned}$

Solving (21.28) and (21.29), we get

$\begin{aligned} x & = y \\ x & = a - y \end{aligned}$

When $x = y$ ,

$\begin{aligned} y & = 0 \\ y & = \frac{a}{3} \end{aligned}$

When $x = a - y$ ,

$\begin{aligned} y & = 0 \\ y & = a \end{aligned}$

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

At $(0, 0)$ ,

$\begin{aligned} V_{x x} & = 0 \\ V_{y y} & = 0 \\ V_{x y} = V_{y x} & = a \\ | \begin{matrix} V_{x x} & V_{x y} \\ V_{y x} & V_{y y} \end{matrix} | = | \begin{matrix} 0 & a \\ a & 0 \end{matrix} | = - a^{2} < 0 \end{aligned}$

At $(a, 0)$ ,

$\begin{aligned} V_{x x} & = 0 \\ V_{y y} & = - 2 a \\ V_{x y} = V_{y x} & = - a \\ | \begin{matrix} V_{x x} & V_{x y} \\ V_{y x} & V_{y y} \end{matrix} | = | \begin{matrix} 0 & - a \\ - a & - 2 a \end{matrix} | = - a^{2} < 0 \end{aligned}$

At $(0, a)$ ,

$\begin{aligned} V_{x x} & = - 2 a < 0 \\ V_{y y} & = 0 \\ V_{x y} = V_{y x} & = - a \\ | \begin{matrix} V_{x x} & V_{x y} \\ V_{y x} & V_{y y} \end{matrix} | = | \begin{matrix} - 2 a & - a \\ - a & 0 \end{matrix} | = - a^{2} < 0 \end{aligned}$

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

At $(\frac{a}{3}, \frac{a}{3})$ ,

$\begin{aligned} V_{x x} & = \frac{- 2 a}{3} \\ V_{y y} & = \frac{- 2 a}{3} \\ V_{x y} = V_{y x} & = \frac{- a}{3} \\ | \begin{matrix} V_{x x} & V_{x y} \\ V_{y x} & V_{y y} \end{matrix} | = | \begin{matrix} \frac{- 2 a}{3} & \frac{- a}{3} \\ \frac{- a}{3} & \frac{- 2 a}{3} \end{matrix} | = \frac{a^{2}}{3} > 0 \end{aligned}$

Thus there is maximum at $(\frac{a}{3}, \frac{a}{3})$ . 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).

$\begin{aligned} d & = [(x - x_{1})^{2} + (y - y_{1})^{2}] + [(x - x_{2})^{2} + (y - y_{2})^{2}] + [(x - x_{3})^{2} + (y - y_{3})^{2}] \end{aligned}$

What we have to do is minimize the $d$ .

$\begin{aligned} d_{x} & = 2 (x - x_{1}) + 2 (x - x_{2}) + 2 (x - x_{3}) \\ d_{y} & = 2 (y - y_{1}) + 2 (y - y_{2}) + 2 (y - y_{3}) \end{aligned}$

For extreme values,

$\begin{aligned} d_{x} & = 0 \\ 2 (x - x_{1}) + 2 (x - x_{2}) + 2 (x - x_{3}) & = 0 \\ x & = \frac{x_{1} + x_{2} + x_{3}}{3} \end{aligned}$

and

$\begin{aligned} d_{y} & = 0 \\ 2 (y - y_{1}) + 2 (y - y_{2}) + 2 (y - y_{3}) & = 0 \\ y & = \frac{y_{1} + y_{2} + y_{3}}{3} \end{aligned}$

Thus $d$ has extreme value at $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

Now calculating partial second derivatives,

$\begin{aligned} d_{x x} & = 6 \\ d_{x y} & = 0 \\ d_{y y} & = 6 \\ d_{y x} & = 0 \end{aligned}$

Thus $d_{x x} = 6 > 0$ and

$| \begin{matrix} d_{x x} & d_{x y} \\ d_{y x} & d_{y y} \end{matrix} | = | \begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix} | = 36 > 0$

Thus $d$ is minimum at $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ . Hence, a point within a triangle such that the sum of the squares of its distances from the three angular points is minimum is $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ .

Differential Calculus

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

21.0.1 Question 5

21.0.2 Question 6

21.0.3 Question 7

21.0.4 Question 8

21.0.5 Question 9

21.1 Useful Links