If x, y and z are real numbers such that: x + y + z = 5 and xy + yz
+ zx = 3. What is the largest value that any of the numbers can
Four rods are hinged at their ends to form a quadrilateral with
fixed side lengths. Show that the quadrilateral has a maximum area
when it is cyclic.
Investigate the graphs of y = [1 + (x - t)^2][1 + (x + t^)2] as the
parameter t varies.
Given that $u > 0$ and $v > 0$ what is the smallest
possible value of $1/u + 1/v$ given that $u + v = 5$?
Here is a solution from Danny of Milliken Mills High School,
Let $S$ be the mininum value,
$S = 1/u + 1/v = (v + u)/(uv)$.
Since $u + v = 5$ then $S = 5/uv$.
The maximum value of $uv$ gives a mininum value of $S$. Given $u
+ v = 5$ then $v = 5 - u$.
Let $f(u)$ give the value of $uv$ as $u$ and $v$ change. Then $
f(u) = u(5 - u) = -u^2 + 5u $. This is a quadratic function and the
vertex $(u, M)$ of the graph of $f(u)$ is at $(2.5, 6.25)$. It
means the maximum value occurs at $M = 6.25$ when $u = 2.5$ and $v
= 5 - u = 2.5$. So the mininum value $S = 1/2.5 + 1/2.5 = 0.8$.
Vassil from Lawnswood High School, Leeds used the same method,
and drew two graphs to illustrate it, one to show the possible
values of $u$ and $v$ where $u + v =5$ and the other, shown below,
the graph of $f(u) = u(5 - u) = -u^2 + 5u$ for the corresponding
range of values of $u$ and $v$.
Peter and Koopa of Boston College used the Arithmetic Mean -
Geometric Mean inequality (AM-GM) noting as above that minimising
$1/u + 1/v$ is the same as maximising $uv$. Here is Peter's
From the AM-GM inequality:
arg = $uv \leq [(u+v)/2]^2 = (5/2)^2 = 25/4.$
Hence the minimum value of $1/u + 1/v$ subject to $u+v = 5$,
$u$, $v > 0$, is $4/5$.