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Question : 30 of 160
Marks:
+1,
-0
Solution:
Let
p=+a+ q=+a ⇒r=a+ Volume of parallelopiped
=[pqr] =p.(q×r) V=|| ⇒V=a3−a+1 Differentiating V with respect to a
=3a2−1 Again, differentiating w.r.t. a
=6a For stationary points
=0 ⇒3a2−1=0 or
a2=‌or‌a=± At,
a= =6.>0 ∴ Volume is minimum at
a=
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