Let `A_(1), A_(2), A_(3), A_(4)` be the areas of the triangular faces of a tetrahedron, and `h_(1), h_(2), h_(3), h_(4)` be the corresponding altitudes of the tetrahedron. If the volume of tetrahedron is `1//6` cubic units, then find the minimum value of ` (A_(1) +A_(2) + A_(3) + A_(4))(h_(1)+ h_(2)+h_(3)+h_(4))` (in cubic units).

Let `A_(1), A_(2), A_(3), A_(4)` be the areas of the triangular faces

1.	Let `A_(1), A_(2), A_(3), A_(4)` be the areas of the triangular faces of a tetrahedron, and `h_(1), h_(2), h_(3), h_(4)` be the corresponding altitudes of the tetrahedron. If the volume of tetrahedron is `1//6` cubic units, then find the minimum value of ` (A_(1) +A_(2) + A_(3) + A_(4))(h_(1)+ h_(2)+h_(3)+h_(4))` (in cubic units).
Answer» Correct Answer - `8` Volume `(V)= (1)/(3) A_(1)h_(1) or h_(1) = (3V)/(A_(1))` Similarly `h_(2)= (3V)/(A_(2)), h_(3)= (3V)/(A_(3)) and h_(4)= (3V)/(A_(4))` So `" "(A_(1)+ A_(2)+ A_(3)+A_(4))(h_(1)+h_(2)+ h_(3)+h_(4))` `" "= (A_(1)+A_(2)+A_(3)+A_(4))((3V)/(A_(1))+ (3V)/(A_2)+ (3V)/(A_3)+ (3V)/(A_4))` `" "= 3V(A_(1)+ A_(2)+A_(3)+A_(4))((1)/(A_1)+ (1)/(A_2)+ (1)/(A_3)+ (1)/(A_4))` Now using A.M.-H.M inequality in `A_(1), A_(2), A_(3), A_(4)`, we get `" "(A_1+A_2+A_3+A_4)/(4) ge (4)/(((1)/(A_1)+ (1)/(A_2)+ (1)/(A_3)+ (1)/(A_4)))` or `" "(A_(1) + A_(2) + A_(3) + A_(4))((1)/(A_1)+ (1)/(A_2)+ (1)/(A_3)+ (1)/(A_4)) ge 16` Hence, the minimum value of `(A_(1)+ A_(2)+ A_(3)+A_(4))(h_(1)+h_(2)+h_(3)+h_(4)) = 3V(16)= 48V= 48(1//6)=8`