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Given an area $A$ that is bounded by the graph of a function $f(x)$ whose only zero is at $x_{0}$ and the x- and y-axis, find a constant $k > 0$, such that the area $A$ is minimal.

$f(x)=-kx+\frac{1}{1-k}$

The only thing I got so far is its area in terms of $k$ and $x_{0}$:

$\int_{0}^{x_{0}} (-kx+\frac{1}{1-k})dx=\left [ -\frac{1}{2}kx^2+\frac{x}{1-k} \right ]^{x_{0}}_{0}=-\frac{1}{2}kx_{0}^2+\frac{x_{0}}{1-k}$

$A(k)=\frac{1}{2}kx_{0}^2+\frac{x_{0}}{1-k}$

At this point I do not know how to continue and seek for your help.

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Hint: Can you find $x_0$ in terms of $k$?

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    I got $\frac{2}{k^{2}(k-1)}$ for $x_{0}$. But this seems to get messy rather quickly when putting that in the function $A(k)$ and then derive. So I was thinking there might be a better way or my result for $x_{0}$ is wrong. – 2017-02-06
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    Observe that $$-kx+\frac{1}{1-k}=0\iff kx=\frac{1}{1-k}\iff x=\frac{1}{k(1-k)}$$ – 2017-02-06