Let $x$ be a variable and $\Phi(\cdot)$ be the CDF of standard Gaussian distribution given by $$\Phi(x)=\int_{-\infty}^x \frac{1}{\sqrt{2\pi} } e^{-\frac{u^2}{2}}du.$$ I want to calculate the maximum of the following function with respect to $x$, $$f(x)=\Phi(x)\Phi(t-kx), $$ where $t$ and $k$ are constants. I have simulated this in MATLAB, and found this function is not concave. However, the simulated figure shows that there is only one maximum.
For example, the figure below shows $f(x)$ vs. $x$ under the case when $t=4,k=3$. The maximum is obtained when $x=0.81$. For arbitrary $t$ and $k$, what is the solution?
