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MA208课程简介
Teacher responsible
Prof Julia Boettcher
Availability
This course is available on the BSc in Actuarial Science, BSc in Business Mathematics and Statistics, BSc in Data Science, BSc in Mathematics and Economics, BSc in Mathematics with Economics and BSc in Mathematics, Statistics and Business. This course is available as an outside option to students on other programmes where regulations permit. This course is available with permission to General Course students.
Prerequisites
Based on the relevant mathematical theory, the course describes various techniques of optimisation and shows how they can be applied. More precisely, the topics covered are: Introduction and review of mathematical background. Introduction to combinatorial optimisation; shortest paths in directed graphs; algorithms and their running time. Classical results on continuous optimisation: Weierstrass’s Theorem concerning continuous functions on compact sets; optimisation of differentiable functions on open sets; Lagrange’s Theorem on equality constrained optimisation; Karush, Kuhn, and Tucker’s Theorem on inequality constrained optimisation. Linear programming and duality theory.
MA208 Optimisation Theory HELP（EXAM HELP， ONLINE TUTOR）
Let $A$ be the arithmetic mean of a finite set of positive numbers, with $x$ the smallest of these numbers, and $y$ the largest. Show that
$$
x y \leq A(x+y-A)
$$
with equality if and only if $x=y=A$.
This is equivalent to showing that
$$
A^2-A(x+y)+x y=(A-x)(A-y) \leq 0
$$
which is true, since $A-x \geq 0$ and $A-y \leq 0$.
To elaborate a bit more, we can start by using the definition of arithmetic mean to express $\$ A \$$ in terms of $\$ \times \$$ and $\$ y \$$ :
$$
A=\frac{x+y}{2}
$$
Substituting this into the inequality we want to prove, we get
$$
x y \leq \frac{x+y}{2}\left(x+y-\frac{x+y}{2}\right)=\frac{(x+y)^2}{4}-\frac{x+y}{2} \cdot \frac{x+y}{2}=\frac{(x-y)^2}{4}
$$
which is equivalent to $\$(x-y)^{\wedge} 2 \backslash$ geq $0 \$$, and therefore always true. Equality occurs if and only if $\$ x=y=A \$$, as we wanted to show.
So, the given inequality holds for any finite set of positive numbers, and is a consequence of the fact that the arithmetic mean lies between the smallest and largest elements of the set.
Minimize the function
$$
f(x)=x^2+\frac{1}{x^2}+4 x+\frac{4}{x},
$$
over positive $x$. Note that the minimum value of $f(x, y)$ cannot be found by a straight-forward application of the AGM Inequality to the four terms taken together. Try to find a way of rewriting $f(x)$, perhaps using more than four terms, so that the AGM Inequality can be applied to all the terms.
The product of the four terms is 16 , whose fourth root is 2 , so the AGM Inequality tells us that $f(x) \geq 8$, with equality if and only if all four terms are equal. But it is not possible for all four terms to be equal. Instead, we can write
$$
f(x)=x^2+\frac{1}{x^2}+x+\frac{1}{x}+x+\frac{1}{x}+x+\frac{1}{x}+x+\frac{1}{x} .
$$
These ten terms have a product of 1 , which tells us that
$$
f(x) \geq 10,
$$
with equality if and only if $x=1$.
To elaborate a bit more, we can start by noticing that the function $\$ f(x) \$$ is a sum of four terms, each of which is symmetric in $\$ x \$$ and $\$ 1 / x \$$ :
$$
f(x)=x^2+\frac{1}{x^2}+4 x+\frac{4}{x}=\left(x+\frac{1}{x}\right)^2+4\left(x+\frac{1}{x}\right)
$$
Now, we can introduce two new variables $\$ y=x+1 / x \$$ and $\$ z=x-1 / x \$$. Note that $\$ y \$$ and $\$ z \$$ are both positive for positive $\$ x \$$, and that $\$ y^{\wedge} 2-z^{\wedge} 2=4 x^{\wedge} 2 \$$. This allows us to express $\$ x \$$ and $\$ 1 / x \$$ in terms of $\$ y \$$ and $\$ z \$$ :
$$
x=\frac{y}{2}+\frac{1}{2} \sqrt{y^2-z^2}, \quad \frac{1}{x}=\frac{y}{2}-\frac{1}{2} \sqrt{y^2-z^2}
$$
Substituting these expressions into $\$ f(x) \$$, we get
$$
f(x)=y^2+z^2+2 y+\frac{1}{2} \sqrt{y^2-z^2}
$$
Now, using the AM-GM inequality on the four terms inside the square root, we have
$$
\sqrt{y^2-z^2} \leq \sqrt{y^2}=y
$$
So, we can further simplify the expression for $\$ \mathrm{f}(\mathrm{x}) \$$ :
$$
f(x) \geq y^2+z^2+2 y+\frac{1}{2} y=\frac{5}{2} y^2+2 y+z^2 \geq \frac{5}{2} y^2+2 y=\frac{1}{2}(y+2)^2+1
$$
Since $\$ y=x+1 / x \backslash$ geq $2 \$$ for positive $\$ x \$$, we have $\$ y+2>$ $4 \$$, and therefore
$$
\frac{1}{2}(y+2)^2+1>\frac{1}{2}(4)^2+1=10
$$
Hence, we have shown that $\$ f(x) \backslash$ geq $10 \$$, with equality if and only if $\$ y=4 \$$ (which implies $\$ z=0 \$$ ) and therefore $\$ x=1 \$$.
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