TANSCHE SMTPC12 Differential Calculus Unit 3

Differential Calculus
(TANSCHE Syllabus)
Madurai Kamaraj University
I- B.Sc., Mathematics, I - Semester

ARO Study Circle

3 Partial Differentiation (Continued)

Syllabus

Homogeneous functions – Partial derivatives of a function of two variables – Maxima and Minima of functions of two variables - Lagrange’s method of undetermined multipliers.
Chapter 8 Sections 1.6, 1.7, Sections 4 and 5.

3.1 Homogeneous Functions

Definition of Homogeneous Functions

A function \(f(x,y)\) is said to be a homogeneous function of degree \(n\) if it satisfies the condition: \[f(tx, ty) = t^n f(x,y)\] for all values of \(t\).

General Form of Homogeneous Functions

A homogeneous function of degree \(n\) in two variables \(x\) and \(y\) can be written as: \[f(x,y) = a_0 x^n + a_1 x^{n-1}y + a_2 x^{n-2}y^2 + \cdots + a_{n-1}xy^{n-1} + a_n y^n\]

Alternate Form

We can write a homogeneous function of degree \(n\) as: \[f(x,y) = x^n \left(a_0 + a_1 \dfrac{y}{x} + a_2 \dfrac{y^2}{x^2} + \cdots + a_n \dfrac{y^n}{x^n}\right)\]

That is, \[\boxed{f(x,y) = x^n F\left(\dfrac{y}{x}\right)}\]

Similarly, a homogeneous function of degree \(n\) consisting of \(m\) variables \(x_1, x_2, \ldots, x_m\) can be written as: \[f(x_1, x_2, \ldots, x_m) = x_1^n F\left(\dfrac{x_2}{x_1}, \dfrac{x_3}{x_1}, \ldots, \dfrac{x_m}{x_1}\right)\]

Euler’s Theorem on Homogeneous Functions

If \(f(x,y)\) is a homogeneous function of degree \(n\) in \(x\) and \(y\), then \[\boxed{x\dfrac{\partial f}{\partial x} + y\dfrac{\partial f}{\partial y} = nf}\]

This is known as Euler’s Theorem for homogeneous functions.

Proof of Euler’s Theorem

Let \(f(x,y)\) be a homogeneous function of degree \(n\). Then, \[f(tx, ty) = t^n f(x,y)\]

Differentiating both sides with respect to \(t\): \[\begin{align*} \dfrac{\partial f}{\partial (tx)} \cdot x + \dfrac{\partial f}{\partial (ty)} \cdot y &= n t^{n-1} f(x,y) \\ x f_x(tx, ty) + y f_y(tx, ty) &= n t^{n-1} f(x,y) \end{align*}\]

Putting \(t = 1\), we get: \[x f_x(x,y) + y f_y(x,y) = n f(x,y)\]

\[\boxed{x\dfrac{\partial f}{\partial x} + y\dfrac{\partial f}{\partial y} = nf}\]

Extension to Three Variables

If \(u = f(x,y,z)\) is a homogeneous function of degree \(n\), then \[\boxed{x\dfrac{\partial f}{\partial x} + y\dfrac{\partial f}{\partial y} + z\dfrac{\partial f}{\partial z} = nf}\]

Illustrative Examples

Example 1.1. Verify Euler’s theorem for \(u = x^3 + y^3 + 3xyz\).

Solution . Given, \[u = x^3 + y^3 + 3xyz\]

Here, each term has degree 3:

• \(x^3\) has degree 3
• \(y^3\) has degree 3
• \(3xyz\) has degree 3 (since \(1+1+1=3\))

Therefore, \(u\) is a homogeneous function of degree \(n = 3\).

Now, find the partial derivatives: \[\begin{align*} \dfrac{\partial u}{\partial x} &= 3x^2 + 3yz \\ \dfrac{\partial u}{\partial y} &= 3y^2 + 3xz \\ \dfrac{\partial u}{\partial z} &= 3z^2 + 3xy \end{align*}\]

Now, compute \(x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z\dfrac{\partial u}{\partial z}\): \[\begin{align*} x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z\dfrac{\partial u}{\partial z} &= x(3x^2 + 3yz) + y(3y^2 + 3xz) + z(3z^2 + 3xy) \\ &= 3x^3 + 3xyz + 3y^3 + 3xyz + 3z^3 + 3xyz \\ &= 3(x^3 + y^3 + z^3 + 3xyz) \\ &= 3u \end{align*}\]

Since \(n = 3\), we have \(x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z\dfrac{\partial u}{\partial z} = 3u\).

Hence, Euler’s theorem is verified.

Example 1.2. If \(u = \tan^{-1}\left(\dfrac{x^3+y^3}{x-y}\right)\), prove that \[x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = \sin 2u\]

Solution . Let \[v = \tan u = \dfrac{x^3+y^3}{x-y}\]

Now, \[\begin{align*} v &= \dfrac{x^3+y^3}{x-y} \\ &= \dfrac{x^3\left(1 + \dfrac{y^3}{x^3}\right)}{x\left(1 - \dfrac{y}{x}\right)} \\ &= x^2 \dfrac{1 + \left(\dfrac{y}{x}\right)^3}{1 - \dfrac{y}{x}} \end{align*}\]

This is of the form \(x^2 F\left(\dfrac{y}{x}\right)\). Hence, \(v\) is a homogeneous function of degree 2.

By Euler’s theorem, \[x\dfrac{\partial v}{\partial x} + y\dfrac{\partial v}{\partial y} = 2v\]

Since \(v = \tan u\), \[\begin{align*} x\dfrac{\partial}{\partial x}(\tan u) + y\dfrac{\partial}{\partial y}(\tan u) &= 2\tan u \\ x\sec^2 u \dfrac{\partial u}{\partial x} + y\sec^2 u \dfrac{\partial u}{\partial y} &= 2\tan u \end{align*}\]

\[\begin{align*} \sec^2 u \left(x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y}\right) &= 2\tan u \\ x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} &= \dfrac{2\tan u}{\sec^2 u} \\ &= 2\tan u \cos^2 u \\ &= 2\sin u \cos u \\ &= \sin 2u \end{align*}\]

\[\boxed{x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = \sin 2u}\]

3.2 Partial Derivatives of a Function of Two Functions

3.2.1 Introduction

Consider the case where \(V\) is a function of two variables \(u\) and \(v\), and \(u\) and \(v\) are themselves functions of \(x\) and \(y\). That is, \[V = F(u,v), \quad u = f(x,y), \quad v = g(x,y)\]

Here, \(x\) and \(y\) are independent variables.

3.2.2 Chain Rule for Function of Two Functions

To find the partial derivatives of \(V\) with respect to \(x\) and \(y\), we use the chain rule: \[\dfrac{\partial V}{\partial x} = \dfrac{\partial V}{\partial u} \dfrac{\partial u}{\partial x} + \dfrac{\partial V}{\partial v} \dfrac{\partial v}{\partial x}\] and \[\dfrac{\partial V}{\partial y} = \dfrac{\partial V}{\partial u} \dfrac{\partial u}{\partial y} + \dfrac{\partial V}{\partial v} \dfrac{\partial v}{\partial y}\]

3.2.3 Operator Notation

We can write the partial derivatives as operators: \[\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x} \dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x} \dfrac{\partial}{\partial v}\] and \[\dfrac{\partial}{\partial y} = \dfrac{\partial u}{\partial y} \dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial y} \dfrac{\partial}{\partial v}\]

3.2.4 Higher Order Partial Derivatives

Using the operator notation, we can find higher order partial derivatives.

3.2.4.1 Second Order Partial Derivative with respect to \(x\)

\[\dfrac{\partial^2 V}{\partial x^2} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial V}{\partial x}\right)\] \[\dfrac{\partial^2 V}{\partial x^2} = \left(\dfrac{\partial u}{\partial x} \dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x} \dfrac{\partial}{\partial v}\right) \left(\dfrac{\partial V}{\partial x}\right)\]

3.2.4.2 Second Order Partial Derivative with respect to \(y\)

\[\dfrac{\partial^2 V}{\partial y^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial V}{\partial y}\right)\] \[\dfrac{\partial^2 V}{\partial y^2} = \left(\dfrac{\partial u}{\partial y} \dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial y} \dfrac{\partial}{\partial v}\right) \left(\dfrac{\partial V}{\partial y}\right)\]

3.2.5 Illustrative Example

Example 1.3. Verify Euler’s Theorem when \(u = x^3 + y^3 + z^3 + 3xyz\).

Solution . Given, \[u = x^3 + y^3 + z^3 + 3xyz\]

Here, each term has degree 3:

• \(x^3\) has degree 3
• \(y^3\) has degree 3
• \(z^3\) has degree 3
• \(3xyz\) has degree 3 (since \(1+1+1=3\))

Therefore, \(u\) is a homogeneous function of degree \(n = 3\).

Now, find the partial derivatives: \[\begin{align*} \dfrac{\partial u}{\partial x} = 3x^2 + 3yz , \qquad \dfrac{\partial u}{\partial y} = 3y^2 + 3zx , \qquad \dfrac{\partial u}{\partial z} = 3z^2 + 3xy \end{align*}\]

Now, compute \(x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z\dfrac{\partial u}{\partial z}\): \[\begin{align*} x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z\dfrac{\partial u}{\partial z} &= x(3x^2 + 3yz) + y(3y^2 + 3zx) + z(3z^2 + 3xy) \\ &= 3x^3 + 3xyz + 3y^3 + 3xyz + 3z^3 + 3xyz \\ &= 3(x^3 + y^3 + z^3 + 3xyz) \\ &= 3u \end{align*}\]

Since \(n = 3\), by Euler’s theorem, we should have: \[x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z\dfrac{\partial u}{\partial z} = 3u\]

Hence, Euler’s theorem is verified.

Example 1.4. If \(u = \tan^{-1}\left(\dfrac{x^3+y^3}{x-y}\right)\), prove that \[x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = \sin 2u\]

Solution . Given, \[u = \tan^{-1}\left(\dfrac{x^3+y^3}{x-y}\right)\]

Let \[v = \tan u = \dfrac{x^3+y^3}{x-y}\]

Now, \[\begin{align*} v &= \dfrac{x^3+y^3}{x-y} \\ &= \dfrac{x^3\left(1 + \dfrac{y^3}{x^3}\right)}{x\left(1 - \dfrac{y}{x}\right)} \\ &= x^2 \dfrac{1 + \left(\dfrac{y}{x}\right)^3}{1 - \dfrac{y}{x}} \end{align*}\]

This is of the form \(x^2 F\left(\dfrac{y}{x}\right)\). Hence, \(v\) is a homogeneous function of degree 2.

By Euler’s theorem, \[x\dfrac{\partial v}{\partial x} + y\dfrac{\partial v}{\partial y} = 2v\]

Since \(v = \tan u\), \[\begin{align*} x\dfrac{\partial}{\partial x}(\tan u) + y\dfrac{\partial}{\partial y}(\tan u) &= 2\tan u \\ x\sec^2 u \dfrac{\partial u}{\partial x} + y\sec^2 u \dfrac{\partial u}{\partial y} &= 2\tan u \\ \sec^2 u \left(x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y}\right) &= 2\tan u \\ x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} &= \dfrac{2\tan u}{\sec^2 u} \\ &= 2\tan u \cos^2 u \\ &= 2\sin u \cos u \\ &= \sin 2u \end{align*}\]

\[\boxed{x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = \sin 2u}\]

3.3 Maxima and Minima of Functions of Two Variables

Just as we studied maxima and minima for functions of a single variable, we can extend these concepts to functions of two variables. A function \(f(x,y)\) can have maximum or minimum values at certain points in its domain.

Definition of Maximum and Minimum

A function \(f(x,y)\) is said to have a maximum value at \((a,b)\) if \[f(a+h, b+k) < f(a,b)\] for all sufficiently small values of \(h\) and \(k\) (whatever their signs may be).

Similarly, \(f(x,y)\) is said to have a minimum value at \((a,b)\) if \[f(a+h, b+k) > f(a,b)\] for all sufficiently small values of \(h\) and \(k\) (whatever their signs may be).

Necessary Conditions for Maxima and Minima

For \(f(x,y)\) to have a maximum or minimum at \((a,b)\), the following conditions must hold: \[\boxed{\left(\dfrac{\partial f}{\partial x}\right)_{a,b} = 0 \quad \text{and} \quad \left(\dfrac{\partial f}{\partial y}\right)_{a,b} = 0}\]

These points are called critical points or stationary points.

Taylor’s Expansion for Two Variables

Using Taylor’s theorem for functions of two variables: \[\begin{align*} f(a+h, b+k) - f(a,b) &= \left(h\dfrac{\partial f}{\partial x} + k\dfrac{\partial f}{\partial y}\right)_{a,b} + \dfrac{1}{2!}\left(h^2\dfrac{\partial^2 f}{\partial x^2} + 2hk\dfrac{\partial^2 f}{\partial x\partial y} + k^2\dfrac{\partial^2 f}{\partial y^2}\right)_{a,b} + \cdots \end{align*}\]

At a critical point, \(\left(\dfrac{\partial f}{\partial x}\right)_{a,b} = \left(\dfrac{\partial f}{\partial y}\right)_{a,b} = 0\).

So, \[f(a+h, b+k) - f(a,b) = \dfrac{1}{2}(h^2 r + 2hk s + k^2 t) + \cdots\]

where \[r = \left(\dfrac{\partial^2 f}{\partial x^2}\right)_{a,b}, \quad s = \left(\dfrac{\partial^2 f}{\partial x\partial y}\right)_{a,b}, \quad t = \left(\dfrac{\partial^2 f}{\partial y^2}\right)_{a,b}\]

Sufficient Conditions for Maxima and Minima

The expression \(h^2 r + 2hk s + k^2 t\) can be written as: \[h^2 r + 2hk s + k^2 t = r\left\{\left(h + \dfrac{ks}{r}\right)^2 + \dfrac{k^2(rt - s^2)}{r^2}\right\}\]

From this, we get the following conditions:

3.3.0.1 Case 1: When \(rt - s^2 > 0\)

If \(rt - s^2 > 0\), then \(h^2 r + 2hk s + k^2 t\) has the same sign as \(r\) for all values of \(h\) and \(k\).

• If \(r > 0\), then \(f(a+h, b+k) - f(a,b) > 0\), so \(f(a,b)\) is a minimum.
• If \(r < 0\), then \(f(a+h, b+k) - f(a,b) < 0\), so \(f(a,b)\) is a maximum.

3.3.0.2 Case 2: When \(rt - s^2 < 0\)

If \(rt - s^2 < 0\), then \(h^2 r + 2hk s + k^2 t\) changes sign for different values of \(h\) and \(k\). Hence, \(f(a,b)\) is neither a maximum nor a minimum. Such points are called saddle points.

3.3.0.3 Case 3: When \(rt - s^2 = 0\)

If \(rt - s^2 = 0\), the test fails. We need further investigation using higher-order derivatives.

3.3.1 Working Rule for Finding Maxima and Minima

To find the maxima and minima of a function \(u = f(x,y)\):

1. Find \(\dfrac{\partial u}{\partial x}\) and \(\dfrac{\partial u}{\partial y}\).
2. Equate them to zero and solve for \(x\) and \(y\) to get critical points \((a_1,b_1), (a_2,b_2), \ldots\)
3. For each critical point, compute: \[r = \dfrac{\partial^2 u}{\partial x^2}, \quad s = \dfrac{\partial^2 u}{\partial x\partial y}, \quad t = \dfrac{\partial^2 u}{\partial y^2}\]
4. Compute \(rt - s^2\):
   • If \(rt - s^2 > 0\) and \(r > 0\) \(\rightarrow\) minimum
   • If \(rt - s^2 > 0\) and \(r < 0\) \(\rightarrow\) maximum
   • If \(rt - s^2 < 0\) \(\rightarrow\) saddle point (no maxima or minima)
   • If \(rt - s^2 = 0\) \(\rightarrow\) test fails (need further investigation)
5. Find the maximum or minimum values by substituting the critical points in \(u = f(x,y)\).

Illustrative Examples

Example 1.5. Find the maximum or minimum values of \[u = 2(x^2 - y^2) - x^4 + y^4\]

Solution . Given, \[u = 2x^2 - 2y^2 - x^4 + y^4\]

Step 1: Find the first partial derivatives. \[\begin{align*} \dfrac{\partial u}{\partial x} &= 4x - 4x^3 = 4x(1 - x^2) \\ \dfrac{\partial u}{\partial y} &= -4y + 4y^3 = 4y(y^2 - 1) \end{align*}\]

Step 2: Equate them to zero. \[\begin{align*} 4x(1 - x^2) &= 0 \implies x = 0, \pm 1 \\ 4y(y^2 - 1) &= 0 \implies y = 0, \pm 1 \end{align*}\]

Critical points are: \[(0,0), (0,1), (0,-1), (1,0), (-1,0), (1,1), (1,-1), (-1,1), (-1,-1)\]

Step 3: Find the second partial derivatives. \[\begin{align*} r = \dfrac{\partial^2 u}{\partial x^2} &= 4 - 12x^2 = 4(1 - 3x^2) \\ s = \dfrac{\partial^2 u}{\partial x\partial y} &= 0 \\ t = \dfrac{\partial^2 u}{\partial y^2} &= -4 + 12y^2 = 4(3y^2 - 1) \end{align*}\]

Step 4: Compute \(rt - s^2\) at each critical point.

\[rt - s^2 = 16(1 - 3x^2)(3y^2 - 1)\]

• At \((0,0)\): \(rt - s^2 = 16(1)(-1) = -16 < 0\) \(\rightarrow\) Saddle point
• At \((0,1)\): \(rt - s^2 = 16(1)(2) = 32 > 0\), \(r = 4 > 0\) \(\rightarrow\) Minimum
• At \((0,-1)\): \(rt - s^2 = 16(1)(2) = 32 > 0\), \(r = 4 > 0\) \(\rightarrow\) Minimum
• At \((1,0)\): \(rt - s^2 = 16(-2)(-1) = 32 > 0\), \(r = -8 < 0\) \(\rightarrow\) Maximum
• At \((-1,0)\): \(rt - s^2 = 16(-2)(-1) = 32 > 0\), \(r = -8 < 0\) \(\rightarrow\) Maximum
• At \((1,1)\): \(rt - s^2 = 16(-2)(2) = -64 < 0\) \(\rightarrow\) Saddle point
• At \((1,-1)\): \(rt - s^2 = 16(-2)(2) = -64 < 0\) \(\rightarrow\) Saddle point
• At \((-1,1)\): \(rt - s^2 = 16(-2)(2) = -64 < 0\) \(\rightarrow\) Saddle point
• At \((-1,-1)\): \(rt - s^2 = 16(-2)(2) = -64 < 0\) \(\rightarrow\) Saddle point

Step 5: Find the maximum and minimum values.

At \((0,1)\) and \((0,-1)\): \[u = 2(0) - 2(1) - 0 + 1 = -1\] Minimum value = \(-1\).

At \((1,0)\) and \((-1,0)\): \[u = 2(1) - 0 - 1 + 0 = 1\] Maximum value = \(1\).

\[\boxed{\text{Maximum value} = 1, \quad \text{Minimum value} = -1}\]

Example 1.6. Discuss the maxima and minima of the function \[u = x^3 y^2 (6 - x - y)\]

Solution . Given, \[u = x^3 y^2 (6 - x - y)\]

Step 1: Find the first partial derivatives. \[\begin{align*} \dfrac{\partial u}{\partial x} &= x^2 y^2 (18 - 4x - 3y) \\ \dfrac{\partial u}{\partial y} &= x^3 y (12 - 2x - 3y) \end{align*}\]

Step 2: Equate them to zero. \[x^2 y^2 (18 - 4x - 3y) = 0 \quad \text{and} \quad x^3 y (12 - 2x - 3y) = 0\]

This gives: \[x = 0 \quad \text{or} \quad y = 0 \quad \text{or} \quad 4x + 3y = 18\] \[\text{and} \quad x = 0 \quad \text{or} \quad y = 0 \quad \text{or} \quad 2x + 3y = 12\]

The non-zero solution is obtained by solving: \[4x + 3y = 18 \quad \text{and} \quad 2x + 3y = 12\]

Subtracting: \(2x = 6 \implies x = 3\), and \(y = 2\).

So the critical point of interest is \((3,2)\).

Step 3: Find the second partial derivatives. \[\begin{align*} r = \dfrac{\partial^2 u}{\partial x^2} &= 6xy^2(6 - 2x - y) \\ s = \dfrac{\partial^2 u}{\partial x\partial y} &= x^2 y(36 - 8x - 9y) \\ t = \dfrac{\partial^2 u}{\partial y^2} &= x^3(12 - 2x - 6y) \end{align*}\]

Step 4: Compute \(rt - s^2\) at \((3,2)\).

\[\begin{align*} r &= 6(3)(2^2)(6 - 6 - 2) = 72(-2) = -144 \\ s &= (3^2)(2)(36 - 24 - 18) = 18(-6) = -108 \\ t &= 27(12 - 6 - 12) = 27(-6) = -162 \end{align*}\]

\[rt - s^2 = (-144)(-162) - (-108)^2 = 23328 - 11664 = 11664 > 0\]

Since \(rt - s^2 > 0\) and \(r < 0\), the function has a maximum at \((3,2)\).

Step 5: Find the maximum value. \[u(3,2) = 3^3 \cdot 2^2 (6 - 3 - 2) = 27 \cdot 4 \cdot 1 = 108\]

\[\boxed{\text{Maximum value} = 108 \text{ at } (3,2)}\]

Example 1.7. Show that, if the perimeter of a triangle is constant, the triangle has a maximum area when it is equilateral.

Solution . Let the sides of the triangle be \(a, b, c\). Given that the perimeter is constant: \[a + b + c = 2K \quad (\text{constant})\]

The area of the triangle is given by Heron’s formula: \[\Delta = \sqrt{K(K-a)(K-b)(K-c)}\]

To maximize \(\Delta\), it is sufficient to maximize \(\Delta^2\): \[\Delta^2 = K(K-a)(K-b)(K-c)\]

Since \(K\) is constant, we maximize: \[f(a,b) = (K-a)(K-b)(K-c)\]

But \(c = 2K - a - b\). So, \[f(a,b) = (K-a)(K-b)(a+b-K)\]

Step 1: Find the partial derivatives. \[\begin{align*} \dfrac{\partial f}{\partial a} &= (K-b)(2K - 2a - b) \\ \dfrac{\partial f}{\partial b} &= (K-a)(2K - a - 2b) \end{align*}\]

Step 2: Equate them to zero. \[(K-b)(2K - 2a - b) = 0\] \[(K-a)(2K - a - 2b) = 0\]

The meaningful solution is: \[2K - 2a - b = 0 \quad \text{and} \quad 2K - a - 2b = 0\]

Solving: \[2a + b = 2K \quad \text{and} \quad a + 2b = 2K\]

Subtracting: \(a - b = 0 \implies a = b\).

Substituting in \(2a + b = 2K\): \[2a + a = 2K \implies 3a = 2K \implies a = \dfrac{2K}{3}\]

Therefore, \(a = b = \dfrac{2K}{3}\) and \(c = \dfrac{2K}{3}\).

Step 3: Find the second partial derivatives. \[\begin{align*} r = \dfrac{\partial^2 f}{\partial a^2} &= -2(K-b) \\ s = \dfrac{\partial^2 f}{\partial a\partial b} &= -3K + 2a + 2b \\ t = \dfrac{\partial^2 f}{\partial b^2} &= -2(K-a) \end{align*}\]

At \(a = b = \dfrac{2K}{3}\): \[\begin{align*} r &= -2\left(K - \dfrac{2K}{3}\right) = -\dfrac{2K}{3} \\ s &= -3K + \dfrac{4K}{3} + \dfrac{4K}{3} = -3K + \dfrac{8K}{3} = -\dfrac{K}{3} \\ t &= -2\left(K - \dfrac{2K}{3}\right) = -\dfrac{2K}{3} \end{align*}\]

\[rt - s^2 = \left(-\dfrac{2K}{3}\right)\left(-\dfrac{2K}{3}\right) - \left(-\dfrac{K}{3}\right)^2 = \dfrac{4K^2}{9} - \dfrac{K^2}{9} = \dfrac{3K^2}{9} = \dfrac{K^2}{3} > 0\]

Since \(rt - s^2 > 0\) and \(r < 0\), the function has a maximum at \(a = b = \dfrac{2K}{3}\).

Therefore, \(a = b = c = \dfrac{2K}{3}\). The triangle is equilateral.

\[\boxed{\text{The triangle has maximum area when it is equilateral.}}\]

3.3.2 Key Points to Remember

1. For maxima and minima of \(f(x,y)\):
   • Necessary conditions: \(\dfrac{\partial f}{\partial x} = 0\) and \(\dfrac{\partial f}{\partial y} = 0\)
   • Sufficient conditions: \(rt - s^2 > 0\)
   • If \(r > 0\) and \(rt - s^2 > 0\) \(\rightarrow\) Minimum
   • If \(r < 0\) and \(rt - s^2 > 0\) \(\rightarrow\) Maximum
   • If \(rt - s^2 < 0\) \(\rightarrow\) Saddle point
   • If \(rt - s^2 = 0\) \(\rightarrow\) Test fails (need further investigation)
2. Critical points are also called stationary points.
3. Saddle points are neither maxima nor minima.

3.4 Lagrange’s Method of Undetermined Multipliers

In many practical problems, we need to find the maximum or minimum of a function subject to certain constraints or conditions. For example, we may want to find the maximum volume of a rectangular box given that its surface area is fixed, or find the minimum cost of production given certain resource constraints.

Theoretically, we could eliminate some variables using the constraint equations and then find the maxima or minima of the resulting function. However, this elimination process is often very difficult or impractical. Lagrange’s method provides an elegant way to solve such problems without eliminating variables.

Statement of the Problem

Suppose we have a function of several variables: \[u = f(x_1, x_2, \ldots, x_m)\] but these variables are connected by \(n\) constraint equations: \[\phi_1(x_1, x_2, \ldots, x_m) = 0, \quad \phi_2(x_1, x_2, \ldots, x_m) = 0, \quad \ldots, \quad \phi_n(x_1, x_2, \ldots, x_m) = 0\] where \(n < m\).

We want to find the maximum or minimum values of \(u\) subject to these constraints.

Lagrange’s Method for One Constraint

To understand the method, let us first consider the case of a function of three variables with one constraint:

Find the maximum or minimum of \[u = f(x,y,z)\] subject to the condition \[\phi(x,y,z) = 0\]

Derivation of the Method

For a function \(u = f(x,y,z)\), the total differential is: \[du = \dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y} dy + \dfrac{\partial f}{\partial z} dz\]

For the constraint \(\phi(x,y,z) = 0\), the total differential is: \[d\phi = \dfrac{\partial \phi}{\partial x} dx + \dfrac{\partial \phi}{\partial y} dy + \dfrac{\partial \phi}{\partial z} dz = 0\]

At a maximum or minimum, \(du = 0\). So, \[\dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y} dy + \dfrac{\partial f}{\partial z} dz = 0\]

We multiply the constraint equation by an undetermined constant \(\lambda\) (called the Lagrange multiplier) and add it to the equation for \(du = 0\): \[\left(\dfrac{\partial f}{\partial x} + \lambda \dfrac{\partial \phi}{\partial x}\right) dx + \left(\dfrac{\partial f}{\partial y} + \lambda \dfrac{\partial \phi}{\partial y}\right) dy + \left(\dfrac{\partial f}{\partial z} + \lambda \dfrac{\partial \phi}{\partial z}\right) dz = 0\]

Since \(dx, dy, dz\) are not all independent (they are related by the constraint), we choose \(\lambda\) such that the coefficients of two of the differentials become zero. This gives: \[\dfrac{\partial f}{\partial x} + \lambda \dfrac{\partial \phi}{\partial x} = 0\] \[\dfrac{\partial f}{\partial y} + \lambda \dfrac{\partial \phi}{\partial y} = 0\] \[\dfrac{\partial f}{\partial z} + \lambda \dfrac{\partial \phi}{\partial z} = 0\]

Working Rule for Lagrange’s Method

To find the maximum or minimum of \(f(x,y,z)\) subject to \(\phi(x,y,z) = 0\):

1. Form the function: \[F(x,y,z) = f(x,y,z) + \lambda \phi(x,y,z)\] where \(\lambda\) is an undetermined constant (Lagrange multiplier).
2. Treat \(x, y, z\) as independent variables and find the partial derivatives: \[\dfrac{\partial F}{\partial x} = 0, \quad \dfrac{\partial F}{\partial y} = 0, \quad \dfrac{\partial F}{\partial z} = 0\]
3. Solve these three equations along with the constraint equation: \[\phi(x,y,z) = 0\] to find the values of \(x, y, z\) and \(\lambda\).
4. The values of \(x, y, z\) obtained give the points where \(f(x,y,z)\) has a maximum or minimum.

Extension to Multiple Constraints

If we have \(m\) variables and \(n\) constraints, we form the function: \[F = f + \lambda_1 \phi_1 + \lambda_2 \phi_2 + \cdots + \lambda_n \phi_n\] where \(\lambda_1, \lambda_2, \ldots, \lambda_n\) are undetermined constants.

Then we solve: \[\dfrac{\partial F}{\partial x_i} = 0 \quad (i = 1, 2, \ldots, m)\] along with the \(n\) constraint equations.

Important Note

Lagrange’s method gives the points where the function may have a maximum or minimum. It does not directly tell us whether the point is a maximum or minimum. We need to use physical or geometrical arguments to determine whether it is a maximum or minimum.

Illustrative Examples

Example 1.8. If \(u = a^3 x^2 + b^3 y^2 + c^3 z^2\) where \(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1\), find the minimum value of \(u\).

Solution . We need to find the minimum of: \[u = a^3 x^2 + b^3 y^2 + c^3 z^2\] subject to: \[\phi(x,y,z) = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} - 1 = 0\]

Form the Lagrange function: \[F = a^3 x^2 + b^3 y^2 + c^3 z^2 + \lambda \left(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} - 1\right)\]

Find the partial derivatives: \[\begin{align*} \dfrac{\partial F}{\partial x} &= 2a^3 x - \dfrac{\lambda}{x^2} = 0 \\ \dfrac{\partial F}{\partial y} &= 2b^3 y - \dfrac{\lambda}{y^2} = 0 \\ \dfrac{\partial F}{\partial z} &= 2c^3 z - \dfrac{\lambda}{z^2} = 0 \end{align*}\]

From these equations: \[\begin{align*} 2a^3 x &= \dfrac{\lambda}{x^2} \implies 2a^3 x^3 = \lambda \\ 2b^3 y^3 &= \lambda \\ 2c^3 z^3 &= \lambda \end{align*}\]

Therefore, \[a^3 x^3 = b^3 y^3 = c^3 z^3\]

\[ax = by = cz = k \quad \text{(say)}\]

So, \[x = \dfrac{k}{a}, \quad y = \dfrac{k}{b}, \quad z = \dfrac{k}{c}\]

Substituting in the constraint: \[\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1\] \[\dfrac{a}{k} + \dfrac{b}{k} + \dfrac{c}{k} = 1\] \[\dfrac{a+b+c}{k} = 1 \implies k = a+b+c\]

Therefore, \[x = \dfrac{a+b+c}{a}, \quad y = \dfrac{a+b+c}{b}, \quad z = \dfrac{a+b+c}{c}\]

The minimum value of \(u\): \[\begin{align*} u &= a^3 x^2 + b^3 y^2 + c^3 z^2 \\ &= a^3 \left(\dfrac{a+b+c}{a}\right)^2 + b^3 \left(\dfrac{a+b+c}{b}\right)^2 + c^3 \left(\dfrac{a+b+c}{c}\right)^2 \\ &= (a+b+c)^2(a+b+c) \\ &= (a+b+c)^3 \end{align*}\]

\[\boxed{\text{Minimum value of } u = (a+b+c)^3}\]

Example 1.9. Find the minimum value of \(x^2 + y^2 + z^2\) subject to \(x + y + z = 3a\).

Solution . We need to find the minimum of: \[u = x^2 + y^2 + z^2\] subject to: \[\phi(x,y,z) = x + y + z - 3a = 0\]

Form the Lagrange function: \[F = x^2 + y^2 + z^2 + \lambda (x + y + z - 3a)\]

Find the partial derivatives: \[\begin{align*} \dfrac{\partial F}{\partial x} &= 2x + \lambda = 0 \\ \dfrac{\partial F}{\partial y} &= 2y + \lambda = 0 \\ \dfrac{\partial F}{\partial z} &= 2z + \lambda = 0 \end{align*}\]

From these equations: \[2x = 2y = 2z = -\lambda\] \[x = y = z = -\dfrac{\lambda}{2}\]

Substituting in the constraint: \[x + y + z = 3a\] \[3x = 3a \implies x = a\]

Therefore, \[x = y = z = a\]

The minimum value: \[u = a^2 + a^2 + a^2 = 3a^2\]

\[\boxed{\text{Minimum value of } u = 3a^2 \text{ at } x = y = z = a}\]

Example 1.10. A tent having the form of a cylinder surmounted by a cone is to contain a given volume. If the canvas required is minimum, show that the altitude of the cone is twice that of the cylinder.

Solution . Let the radius of the cylinder be \(x\), the height of the cylinder be \(y\), and the height of the cone be \(z\).

Volume of the tent: \[V = \pi x^2 y + \dfrac{1}{3} \pi x^2 z = \pi K^3 \quad \text{(given constant)}\]

So, \[\pi x^2 y + \dfrac{1}{3} \pi x^2 z = \pi K^3\] \[3x^2 y + x^2 z = 3K^3\]

Surface area of the tent (canvas required): \[S = 2\pi xy + \pi x \sqrt{x^2 + z^2}\]

Since \(\pi\) is constant, we minimize: \[f(x,y,z) = 2xy + x\sqrt{x^2 + z^2}\] subject to: \[\phi(x,y,z) = 3x^2 y + x^2 z - 3K^3 = 0\]

Form the Lagrange function: \[F = 2xy + x\sqrt{x^2 + z^2} + \lambda (3x^2 y + x^2 z - 3K^3)\]

Find the partial derivatives: \[\begin{align*} \dfrac{\partial F}{\partial x} &= 2y + \sqrt{x^2 + z^2} + \dfrac{x^2}{\sqrt{x^2 + z^2}} + \lambda (6xy + 2xz) = 0 \quad (1) \\ \dfrac{\partial F}{\partial y} &= 2x + 3\lambda x^2 = 0 \quad (2) \\ \dfrac{\partial F}{\partial z} &= \dfrac{xz}{\sqrt{x^2 + z^2}} + \lambda x^2 = 0 \quad (3) \end{align*}\]

From equation (2): \[2x + 3\lambda x^2 = 0\] \[x(2 + 3\lambda x) = 0\]

Since \(x \neq 0\) (radius cannot be zero): \[2 + 3\lambda x = 0 \implies \lambda x = -\dfrac{2}{3} \quad (4)\]

From equation (3): \[\dfrac{xz}{\sqrt{x^2 + z^2}} + \lambda x^2 = 0\] \[x\left(\dfrac{z}{\sqrt{x^2 + z^2}} + \lambda x\right) = 0\] \[\dfrac{z}{\sqrt{x^2 + z^2}} + \lambda x = 0\] \[\dfrac{z}{\sqrt{x^2 + z^2}} = -\lambda x = \dfrac{2}{3} \quad \text{(from (4))}\]

Squaring: \[\dfrac{z^2}{x^2 + z^2} = \dfrac{4}{9}\] \[9z^2 = 4x^2 + 4z^2\] \[5z^2 = 4x^2 \implies z^2 = \dfrac{4x^2}{5}\] \[z = \dfrac{2x}{\sqrt{5}} \quad (5)\]

Now, \[\sqrt{x^2 + z^2} = \sqrt{x^2 + \dfrac{4x^2}{5}} = \sqrt{\dfrac{9x^2}{5}} = \dfrac{3x}{\sqrt{5}}\]

Substituting in equation (1): \[\begin{align*} 2y + \dfrac{3x}{\sqrt{5}} + \dfrac{x^2}{\dfrac{3x}{\sqrt{5}}} + \lambda (6xy + 2xz) &= 0 \\ 2y + \dfrac{3x}{\sqrt{5}} + \dfrac{\sqrt{5}x}{3} + \lambda x(6y + 2z) &= 0 \end{align*}\]

Using \(\lambda x = -\dfrac{2}{3}\): \[\begin{align*} 2y + \dfrac{3x}{\sqrt{5}} + \dfrac{\sqrt{5}x}{3} - \dfrac{2}{3}(6y + 2z) &= 0 \\ 2y + \dfrac{3x}{\sqrt{5}} + \dfrac{\sqrt{5}x}{3} - 4y - \dfrac{4z}{3} &= 0 \\ -2y + \dfrac{3x}{\sqrt{5}} + \dfrac{\sqrt{5}x}{3} - \dfrac{4z}{3} &= 0 \end{align*}\]

Substituting \(z = \dfrac{2x}{\sqrt{5}}\): \[\begin{align*} -2y + \dfrac{3x}{\sqrt{5}} + \dfrac{\sqrt{5}x}{3} - \dfrac{8x}{3\sqrt{5}} &= 0 \\ -2y + x\left(\dfrac{3}{\sqrt{5}} + \dfrac{\sqrt{5}}{3} - \dfrac{8}{3\sqrt{5}}\right) &= 0 \\ -2y + x\left(\dfrac{9}{3\sqrt{5}} + \dfrac{5}{3\sqrt{5}} - \dfrac{8}{3\sqrt{5}}\right) &= 0 \\ -2y + x\left(\dfrac{6}{3\sqrt{5}}\right) &= 0 \\ -2y + \dfrac{2x}{\sqrt{5}} &= 0 \\ y = \dfrac{x}{\sqrt{5}} \end{align*}\]

From (5), \(z = \dfrac{2x}{\sqrt{5}}\).

Therefore, \[z = 2y\]

\[\boxed{\text{The height of the cone is twice the height of the cylinder.}}\]

3.4.1 Key Points to Remember

1. Lagrange’s method is used to find maxima/minima of functions subject to constraints.
2. Form the function \(F = f + \lambda \phi\) where \(\lambda\) is the Lagrange multiplier.
3. Solve \(\dfrac{\partial F}{\partial x_i} = 0\) along with the constraint equations.
4. The method gives stationary points, but does not distinguish between maxima and minima.
5. For multiple constraints, use multiple Lagrange multipliers.
6. The method avoids the difficult process of eliminating variables.