Field Optimization

The position vectors of the two particles and test particle are: $\vec{r}_1 = -\hat{i} + y_1 \hat{j}$ $\vec{r}_2 = \hat{i} + y_2 \hat{j}$ $\vec{r}_0 = \hat{j}$

The forces acting on the test particle due to each of the other two particles are: $\vec{F}_{10} = \frac{1}{\mid \vec{r}_1 - \vec{r}_0 \mid^3} \left(\vec{r}_1 - \vec{r}_0\right)$ $\vec{F}_{20} = \frac{1}{\mid \vec{r}_2 - \vec{r}_0 \mid^3} \left(\vec{r}_2 - \vec{r}_0\right)$

The resultant force on the test particle is: $\vec{F}_{t} = \vec{F}_{10} + \vec{F}_{20}$

Its magnitude is: $F_m = \mid \vec{F}_{t} \mid$

The resulting expression which is a function of $y_1$ and $y_2$ is a highly nonlinear expression. The objective of this question is to maximise $F_m$ subject to the constraints imposed on $y_1$ and $y_2$ . One can consider computing partial derivatives and equating it to zero. This method does not account for the imposed constraints on the independent variables ( $y_1$ and $y_2$ ). Solving such an optimization problem would need the satisfaction of the optimality conditions (KKT conditions). Furthermore, the risk here is that there is a possibility of multiple local optima. The occurrence of multiple local optimal points is something I have not checked.

The quickest way is to use a computer program and vary the independent variables in the specified range in small steps until the maximum is found. The answer comes out to be: $\boxed{0.8109}$ .