Given the differential equation: d$^2$T$/$dx$^2=16-($x$-4)^2$ with boundary conditions T$(0)=300$ and T$(8)=400.$ Use the finite difference technique with spacing between the nodes of $$x$=2$ to solve for T between $0$ and $8.$ What is the minimum value in this $\{$T$\}$ vector? a$.238\mathrm{.}4192$ b$.138\mathrm{.}0000$ c$.238\mathrm{.}0000$ d$.170\mathrm{.}0000$