The value of \(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)\) is

(D)
Point of intersection of \(x=4\) and \(3 x+2 y=18\) is \(Q \equiv(4,3)\)
Point of intersection of \(y=6\) and \(3 x+2 y=18\) is \(P \equiv(2,6)\)
Point \(D \equiv(4,0)\) and \(C \equiv(0,6)\) are as shown.
The feasible region of th given L.P.P. is shaded portion CPQ D O.
We have to maximize \(\mathrm{Z}=3 \mathrm{x}+5 \mathrm{y}\).
Now,
\(\begin{array}{ll}\mathrm{Z} \text { at } \mathrm{C}(0,6) & =3(0)+5(6)=30 \\ \mathrm{Z} \text { at } \mathrm{P}(2,6) & =3(2)+5(6)=36 \\ \mathrm{Z} \text { at } \mathrm{Q}(4,3) & =3(4)+5(3)=27 \\ \mathrm{Z} \text { at } \mathrm{D}(4,0) & =3(4)+5(0)=12 \\ \mathrm{Z} \text { at } \mathrm{O}(0,0) & =3(0)+5(0)=0\end{array}\)
Clearly the maximum value of \(Z\) is 36 at \(P(2,6)\)