Let $f ( x , y , z ) = x y z$ with constraint $2 x y + 3 x z + y z = 72$ . Then f has a relative maximum at

A) $( - 2 , - 6 , - 4 )$
B) $( 2 , - 6 , - 4 )$
C) $( - 2 , - 6,4 )$
D) $( 2,6,4 )$
E) $( - 2,6 , - 4 )$

Question

Let $f ( x , y , z ) = x y z$ with constraint $2 x y + 3 x z + y z = 72$ . Then f has a relative maximum at

A) $( - 2 , - 6 , - 4 )$
B) $( 2 , - 6 , - 4 )$
C) $( - 2 , - 6,4 )$
D) $( 2,6,4 )$
E) $( - 2,6 , - 4 )$

Quizplus · Accepted Answer

To find the relative maximum of $f(x, y, z) = xyz$ subject to the constraint $2xy + 3xz + yz = 72$, we can use the method of Lagrange multipliers. This method involves finding the points where the gradient of $f$ is parallel to the gradient of the constraint function $g(x, y, z) = 2xy + 3xz + yz - 72$. The Lagrange function is $L(x, y, z, \lambda) = xyz - \lambda(2xy + 3xz + yz - 72)$.Taking partial derivatives of $L$ with respect to $x$, $y$, $z$, and $\lambda$, and setting them equal to zero gives a system of equations. Solving this system can be complex, but intuitively, for $f(x, y, z) = xyz$ to be maximized under the given constraint, $x$, $y$, and $z$ should be positive because the product of three positive numbers is larger than the product of any combination of positive and negative numbers that satisfy the same linear constraint. Among the given options, only $(2, 6, 4)$ consists of all positive numbers, making it the most likely candidate for a relative maximum.

Let f(x,y,z)=xyzf ( x , y , z ) = x y zf(x,y,z)=xyz

Let $f ( x , y , z ) = x y z$