The maximum volume of a parallelopiped (in cubic units) with vectors (2a i + k ), (a j - a k ) , and (3 i + a j ) , where a [0, 1] , as its coterminous edges is...

Step-by-step solution

The volume V of the parallelopiped is given by the absolute value of the scalar triple product of its coterminous edges. V = | vmatrix 2a & 0 & 1 0 & a & -a 3 & a & 0 vmatrix | Expanding the determinant along the first row: V = |2a(0 - (-a^2)) - 0 + 1(0 - 3a)| V = |2a^3 - 3a| Since a [0, 1] , 2a^3 3a , so V(a) = 3a - 2a^3 . To find the maximum volume, we differentiate V(a) with respect to a and equate it to zero: V'(a) = 3 - 6a^2 = 0 a^2 = 1 2 a = 1 2 (since a [0, 1] ) To verify it is a maximum, we check the second