How small can this triangle be?

How small can this triangle be?

Question

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

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Can you express the area of the triangle as a function of the gradient, $m$?

How can we find the minimum value of a function?

Solution

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.

Approach 1 - Differentiation

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 $m=-2$.

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?

Approach 2 - Two numbers multiplying to a constant If two negative numbers have a constant product, as $\dfrac{m}{2}$ and $\dfrac{2}{m}$ do (i.e. $1$), then their sum is maximised when they are equal.

Can you see why? Note we're being asked to minimise $2-\dfrac{m}{2}-\dfrac{2}{m}$, or in other words to maximise $\dfrac{m}{2}+\dfrac{2}{m}$ remembering that our $m$ and therefore this sum is negative. Imagine that we have two positive numbers that can vary. -- If their product is constant, when is their sum a minimum? -- If their sum is constant, when is their product a maximum? These two questions crop up surprisingly often in mathematics, and the answer to both questions is the same - when the two numbers are equal. Have a think about this: you might like to make sketches to help. What happens if we fix $j$ and let $k$ vary, or if we fix $k$ and let $j$ vary? What can we say about the coordinates of the points where the curves meet?

This is when $\dfrac{m}{2}=\dfrac{2}{m}$, which gives $m = \pm 2$, as before. We need $m$ to be negative, so $m = -2$. Thus using either approach the least possible area of the triangle is $A(-2) = \frac{-(-2)^2 + 4\times(-2) - 4}{2\times(-2)} = \frac{-4 - 8 - 4}{-4} = 4.$

It is interesting to note that in both approaches we get an option of $m=2$ which we discard because $m$ must be negative. We can easily see that when $m=2$ we are on the branch of the curve of $A$ against $m$ in the 4th quadrant at its maximum point. This local maximum is smaller than the local minimum when $m=-2$, it would be the area where a straight line through the point $(1, 2)$ made no triangle at all in any quadrant, which might be seen as a minimum area of $0$!