Introduction
Lagrange Multiplier Method – Solved Problems
The method of Lagrange multipliers helps us find extrema of a function subject to constraints. Below are 5 solved examples.
\[\begin{align}
& F=8xyz+\lambda \left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}+\frac{{{z}^{2}}}{{{c}^{2}}}-1 \right) \\
& \Rightarrow \frac{\partial F}{\partial x}=8yz+\lambda \frac{2x}{{{a}^{2}}}=0 \\
& \Rightarrow 8yz+\lambda \frac{2x}{{{a}^{2}}}=0 \\
& \text{similarily} \\
& 8xz+\lambda \frac{2y}{{{a}^{2}}}=0 \\
& 8xy+\lambda \frac{2z}{{{a}^{2}}}=0 \\
& \Rightarrow \left( 8xyz+\lambda \frac{2{{x}^{2}}}{{{a}^{2}}} \right)+\left( 8xyz+\lambda \frac{2{{y}^{2}}}{{{b}^{2}}} \right)+\left( 8xyz+\lambda \frac{2{{z}^{2}}}{{{c}^{2}}} \right)=0 \\
& \Rightarrow 24xyz+2\lambda \left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}+\frac{{{z}^{2}}}{{{c}^{2}}} \right)=0 \\
& \Rightarrow 24xyz+2\lambda \left( 1 \right)=0 \\
& \Rightarrow \lambda =-12xyz \\
& \Rightarrow 8xz+\lambda \frac{2y}{{{a}^{2}}}=0 \\
& \Rightarrow 8xz-12xyz\times \frac{2y}{{{a}^{2}}}=0 \\
& \Rightarrow 8xz\left( 1-\frac{3{{y}^{2}}}{{{a}^{2}}} \right)=0 \\
& \Rightarrow y=\frac{a}{\sqrt{3}} \\
& \text{similarily, }x=\frac{b}{\sqrt{3}},z=\frac{c}{\sqrt{3}} \\
& \Rightarrow {{V}_{\max }}=8xyz=\frac{8abc}{3\sqrt{3}} \\
\end{align}\]
Problem 1
Maximize \(f(x,y) = xy\) subject to \(x^2 + y^2 = 1\).Solution:
We form \(L(x,y,\lambda) = xy + \lambda(1 - x^2 - y^2)\).
Conditions:
\[ \frac{\partial L}{\partial x} = y - 2\lambda x = 0, \quad \frac{\partial L}{\partial y} = x - 2\lambda y = 0, \quad x^2+y^2=1. \] Solving gives \(x = \pm \tfrac{1}{\sqrt{2}}, \; y = \pm \tfrac{1}{\sqrt{2}}\).Thus, maximum value = \(\tfrac{1}{2}\), minimum = \(-\tfrac{1}{2}\).
Problem 2
Minimize \(f(x,y) = x^2 + y^2\) subject to \(x + y = 1\).Solution:
\(L(x,y,\lambda) = x^2 + y^2 + \lambda(1 - x - y).\) Conditions: \[ \frac{\partial L}{\partial x} = 2x - \lambda = 0, \quad \frac{\partial L}{\partial y} = 2y - \lambda = 0, \quad x+y=1. \] From first two: \(x=y\). Then \(x+y=1 \implies x=y=\tfrac{1}{2}\). Minimum value = \(\tfrac{1}{2}\).Problem 3
Maximize \(f(x,y) = x + y\) subject to \(x^2 + y^2 = 4\).Solution:
\(L = x+y + \lambda(4 - x^2 - y^2).\) Conditions: \[ \frac{\partial L}{\partial x} = 1 - 2\lambda x = 0, \quad \frac{\partial L}{\partial y} = 1 - 2\lambda y = 0. \] So \(x=y\). Constraint: \(2x^2=4 \implies x=\pm \sqrt{2}, y=\pm \sqrt{2}\). Values: \(x+y=2\sqrt{2}\) (maximum), \(-2\sqrt{2}\) (minimum).Problem 4
Minimize \(f(x,y,z) = x^2+y^2+z^2\) subject to \(x+y+z=3\).Solution:
\(L = x^2+y^2+z^2 + \lambda(3 - x - y - z).\) Conditions: \[ 2x-\lambda=0, \quad 2y-\lambda=0, \quad 2z-\lambda=0. \] So \(x=y=z=\tfrac{\lambda}{2}\). Constraint: \(3\cdot \tfrac{\lambda}{2}=3 \implies \lambda=2 \implies x=y=z=1.\) Minimum value = \(1^2+1^2+1^2=3\).Problem 5
Maximize \(f(x,y) = x^2y\) subject to \(x^2+y^2=3\).Solution:
\(L = x^2y + \lambda(3 - x^2 - y^2).\) Conditions: \[ \frac{\partial L}{\partial x} = 2xy - 2\lambda x = 0, \quad \frac{\partial L}{\partial y} = x^2 - 2\lambda y = 0. \] Case 1: \(x=0 \implies y=\pm \sqrt{3}, f=0.\) Case 2: \(y=\lambda, \, x^2 = 2\lambda^2 \implies 3\lambda^2=3 \implies \lambda=\pm 1, \; x^2=2, \, y=\pm1.\) Thus, extrema: \(f=\pm 2.\) Maximum = \(2\), Minimum = \(-2\).Maximize \( xyz \) subject to \( x+y+z=1 \)
\[\begin{align} & F=f\left( x,y \right)+\lambda g\left( x,y \right) \\ & \Rightarrow F=xyz+\left( x+y+z-1 \right) \\ & \Rightarrow \frac{\partial F}{\partial x}=\frac{\partial f\left( x,y \right)}{\partial x}+\lambda \frac{\partial g\left( x,y \right)}{\partial x}=0 \\ & \Rightarrow yz+\lambda =0 \\ & \because \frac{\partial F}{\partial y}=\frac{\partial f\left( x,y \right)}{\partial y}+\lambda \frac{\partial g\left( x,y \right)}{\partial y}=0 \\ & \Rightarrow xz+\lambda =0 \\ & \because \frac{\partial F}{\partial y}=\frac{\partial f\left( x,y \right)}{\partial y}+\lambda \frac{\partial g\left( x,y \right)}{\partial y}=0\, \\ & \Rightarrow xy+\lambda =0 \\ & \Rightarrow \left( xyz+\lambda x \right)+\left( xyz+\lambda y \right)+\left( xyz+\lambda z \right)=0 \\ & \Rightarrow 3xyz+\lambda \left( x+y+z \right)\,=0 \\ & \Rightarrow 3xyz+\lambda \,=0 \\ & \Rightarrow \lambda =-3xyz \\ & \text{Now putting }\lambda \text{ } \\ & xy-3xyz=0 \\ & \Rightarrow 1=3z\Rightarrow z=\frac{1}{3} \\ & \text{similarily x=z=}\frac{1}{3} \\ & \Rightarrow {{\left( xyz \right)}_{\max }}=\frac{1}{27} \\ \end{align}\]Minimize \({x}^{2}}+{{y}^{2}}+{{z}^{2}\) subject to \( x+y+z=1\)
\[\begin{align} & F=f\left( x,y \right)+\lambda g\left( x,y \right) \\ & \Rightarrow F={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+\left( x+y+z-3 \right) \\ & \Rightarrow \frac{\partial F}{\partial x}=\frac{\partial f\left( x,y \right)}{\partial x}+\lambda \frac{\partial g\left( x,y \right)}{\partial x}=0 \\ & \Rightarrow 2x+\lambda =0 \\ & \because \frac{\partial F}{\partial y}=\frac{\partial f\left( x,y \right)}{\partial y}+\lambda \frac{\partial g\left( x,y \right)}{\partial y}=0 \\ & \Rightarrow 2y+\lambda =0 \\ & \because \frac{\partial F}{\partial y}=\frac{\partial f\left( x,y \right)}{\partial y}+\lambda \frac{\partial g\left( x,y \right)}{\partial y}=0\, \\ & \Rightarrow 2z+\lambda =0 \\ & \Rightarrow x=y=z=-\frac{\lambda }{2} \\ & \text{From }x+y+z=3 \\ & -3\frac{\lambda }{2}=3 \\ & \Rightarrow \lambda =-2 \\ & \Rightarrow x=y=z=1 \\ & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}_{\text{maximum}}}=3 \\ \end{align}\]Series Expansions for Undergraduate Students 🎓
I. Taylor Series
The **Taylor Series** is a representation of a function as an infinite sum of terms calculated from the function's derivatives at a single point. It is a fundamental concept for approximating functions.
Definition
If a function $f(x)$ has derivatives of all orders at a point $x=a$, the Taylor series for $f(x)$ centered at $x=a$ is given by:
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$
Expanded form:
$$f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots$$
- $f^{(n)}(a)$ is the $n$-th derivative of $f(x)$ evaluated at $a$.
- The expansion is a polynomial approximation, becoming more accurate as more terms are included.
Maclaurin Series (Taylor Series at $a=0$)
A Taylor series centered at $a=0$ is called a **Maclaurin Series**:
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \dots$$
II. Binomial Series
The **Binomial Series** is a power series expansion for functions of the form $(1+x)^k$, where $k$ is any real number.
Definition
The binomial series expansion is:
$$(1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n$$
Expanded form:
$$1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \dots$$
Where the **generalized binomial coefficient** $\binom{k}{n}$ is defined as:
$$\binom{k}{n} = \frac{k(k-1)(k-2)\dots(k-n+1)}{n!}$$
Radius of Convergence
- If $k$ is a non-negative integer, the series terminates (the usual Binomial Theorem) and is valid for all $x$.
- If $k$ is **not** a non-negative integer, the series is infinite and converges for **$|x| < 1$**.
Example: For $\frac{1}{\sqrt{1+x}} = (1+x)^{-1/2}$, $k = -1/2$. The expansion is valid for $|x| < 1$.
III. Common Maclaurin Series Expansions
These are the essential Maclaurin series derived from the Taylor series formula with $a=0$.
| Function $f(x)$ | Maclaurin Series Expansion | Interval of Convergence |
|---|---|---|
| **Exponential** $e^x$ | $$\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$ | $(-\infty, \infty)$ |
| **Sine** $\sin(x)$ | $$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$$ | $(-\infty, \infty)$ |
| **Cosine** $\cos(x)$ | $$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$ | $(-\infty, \infty)$ |
| **Geometric** $\frac{1}{1-x}$ | $$\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots$$ | $(-1, 1)$ |
| **Logarithm** $\ln(1+x)$ | $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$$ | $(-1, 1]$ |
| **Arctangent** $\arctan(x)$ | $$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$$ | $[-1, 1]$ |
IV. Techniques for Finding New Series
Instead of calculating derivatives, use these algebraic methods to derive series for related functions:
1. Substitution (Composition)
Replace $x$ in a known series with another expression, $g(x)$.
Example: To find the series for $e^{-x^2}$, substitute $u = -x^2$ into the $e^u$ series ($e^u = \sum u^n/n!$):
$$e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \dots = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \dots$$
2. Differentiation
Differentiate the series term-by-term to find the series for the derivative of the function.
Example: Find the series for $\frac{1}{(1-x)^2}$ by differentiating the geometric series $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$:
$$\frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{d}{dx}\left(1 + x + x^2 + x^3 + \dots\right)$$
$$\frac{1}{(1-x)^2} = 0 + 1 + 2x + 3x^2 + \dots = \sum_{n=1}^{\infty} nx^{n-1}$$
3. Integration
Integrate the series term-by-term to find the series for the integral of the function (don't forget the constant of integration, $C$).
Example: The series for $\arctan(x)$ is found by integrating the series for $\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \dots$ (a substitution into the geometric series):
$$\arctan(x) = \int \left(1 - x^2 + x^4 - x^6 + \dots\right) dx = C + x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots$$
Since $\arctan(0)=0$, the constant $C=0$.