Review question

# How small can this triangle be? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7720

## Solution

Write down the equation of the line of gradient $m$ through the point $(1, 2)$. Find, in terms of $m$, the intercepts which this line makes on the two axes.

We can write the equation as $y-2=m(x-1)$ or $y=mx+2-m.$

The $y$-intercept is at $y=2-m$.

The $x$-intercept is given by $0 = mx + 2 - m$, which yields $x = 1-\dfrac{2}{m}$.

If $m = 0$, the line is the horizontal line $y = 2$, so there’s no $x$-intercept.

A line is drawn through the point $(1, 2)$ to form with the axes a triangle in the positive quadrant. Find the least possible area of the triangle.

Note that to form a triangle in the first quadrant, $m$ must be negative.

The area of the blue triangle is $A=\frac{1}{2}(2 - m)\left(1-\frac{2}{m}\right) = \frac{1}{2}\left(-m + 4 - \frac{4}{m}\right).$

So the least area of the triangle is given by the minimum of the function $A(m) = 2-\dfrac{m}{2}-\dfrac{2}{m}$.

We can sketch this graph by adding the curves $y=2$, $y= -\dfrac{x}{2}$ and $y = -\dfrac{2}{x}$.

For a triangle to be created, $m$ must be negative, so we’re looking for the minimum that we can see on the graph.

We can find the minimum by differentiating $A$. We have $\dfrac{dA}{dm}= -\dfrac{1}{2}+\dfrac{2}{m^2}.$

Therefore $\dfrac{dA}{dm}=0$ when $m=\pm 2$. Since $m$ has to be negative, $m = -2$.

To check if this is a minimum point consider the gradients either side of $m=-2$.

$m$ $-2.1$ $-2$ $-1.9$
$\dfrac{dA}{dm}$ $-\dfrac{1}{2}+\dfrac{2}{2.1^2}<0$ $0$ $-\dfrac{1}{2}+\dfrac{2}{1.9^2}>0$

So the least possible area of the triangle is $A(-2) = 2-\frac{-2}{2}-\frac{2}{-2} = 4.$

As an alternative to using calculus, we could notice that the two terms $\dfrac{m}{2}$ and $\dfrac{2}{m}$ multiply to one and that we are looking for the maximum of their sum.

This is a situation that crops up surprisingly often in mathematics and has a simple solution. It turns out that the maximum of $\dfrac{m}{2}+\dfrac{2}{m}$ occurs when the two terms are equal.

To see why this is, you might think about the symmetry or you might think about the graphs of $xy=1$ and $x+y=k$.

Using this fact leads to the solution $m=-2$ as above.

The graph of $A(m)$ also has a stationary point at $m=+2$. What does this correspond to geometrically?