Loaded die parameter maximising a total of seven

A six-sided die is weighted so that faces $1$ through $6$ come up with probabilities $\tfrac16$, $\tfrac15-x$, $\tfrac1{10}+2x$, $\tfrac15-x$, $\tfrac16$ and $\tfrac16$ respectively. You roll it twice. Which value of $x$ makes a total of $7$ most likely?

$x=1/30$. A total of $7$ comes from the unordered pairs $\{1,6\},\{2,5\},\{3,4\}$, each counted twice for order. This gives $P(x)=\tfrac1{18}+\tfrac13\left(\tfrac15-x\right)+2\left(\tfrac1{10}+2x\right)\left(\tfrac15-x\right)=-4x^2+\tfrac{4}{15}x+\text{const}$, maximised at $x=\tfrac{1}{30}$.