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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
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The correct answer is:
$\frac{\pi}{4}$
(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 $P(2,6)$
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 $P(2,6)$
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