What's the minimum area of this variable triangle?

What's the minimum area of this variable triangle?

Question

$OABC$ is a square of side $1\,\text{ft.}$. A point $X$ in $AB$ and a point $Y$ in $BC$ are such that $AX=x\,\text{ft.}$, $BY=kx\,\text{ft.}$ and $BY$ is longer than $AX$. For a given value of $k$, show that the minimum area of the triangle $OXY$ is $\frac{4k-1}{8k}\,\text{sq.ft.}$

Suggestion

Could we draw a picture of the problem?

Can we write the area of the triangle $OXY$ as a function of $x$?

Interactive diagram

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/j46H5ZGN/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

The bar labelled ‘Area’ has a height that measures the area of triangle $OXY$.

The bar labelled ‘Bound’ shows the value of $\dfrac{4k-1}{8k}$.

For a given $k$, how does the area vary with $x$?

Solution

Don’t get distracted by the choice of feet for units here, the question works perfectly well whatever the units given for length.

From the picture, we can see that the easiest way to find the area of the triangle $OXY$ is to subtract the areas of the triangles $OAX$, $BXY$ and $OCY$ from the area of the square $OABC$.

Writing the area as a function of $x$, $A(x)$, we find that $A(x)=1-\frac{x}{2}-\frac{1}{2}kx(1-x)-\frac{1}{2}(1-kx)=\frac{1}{2}-\frac{x}{2}+\frac{kx^2}{2}.$

Taking the derivative with respect to $x$, we find that $\frac{dA}{dx}=-\frac{1}{2}+kx,$ which is equal to zero when $x=\dfrac{1}{2k}$. Since $A(x)$ has a positive $x^2$ coefficient, this must be a ‘vertex-down’ parabola, and we have a minimum.

Alternatively, since $A(x)$ is a quadratic, we could have found the minimum point by completing the square.

So we have a minimum value of $A\left(\dfrac{1}{2k}\right)=\frac{1}{2}-\frac{1}{4k}+\frac{1}{8k}=\frac{1}{2}-\frac{1}{8k}=\frac{4k-1}{8k}.$

Therefore, for fixed $k$, the minimum area of the triangle $OXY$ is $\frac{4k-1}{8k}\,\text{sq.ft.}$

How does this expression behave as $k$ varies?