Let, ƒ(x,y,z) = 4x²+7xy+3xz² .The direction in which the function ƒ(x,y,z) increases most rapidly at point P= (1,0,2) is

To find the direction in which a scalar function $f(x,y,z)$ increases most rapidly at a given point, we need to calculate the gradient of the function, $\nabla f$, at that point.

The gradient vector of $f(x,y,z)$ is given by:
$\nabla f=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$

The given function is:
$f(x,y,z)=4x^2+7xy+3x{z}^2$

Let us compute the first-order partial derivatives:
1. With respect to $x$:
$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}(4x^2+7xy+3x{z}^2)=8x+7y+3z^2$
2. With respect to $y$:
$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}(4x^2+7xy+3x{z}^2)=7x$
3. With respect to $z$:
$\frac{\partial f}{\partial z}=\frac{\partial }{\partial z}(4x^2+7xy+3x{z}^2)=6xz$

Now, we evaluate these partial derivatives at the given point $P=(1,0,2)$, where $x=1$, $y=0$, and $z=2$:
${\left(\frac{\partial f}{\partial x}\right)}_{(1,0,2)}=8\left(1\right)+7\left(0\right)+3{\left(2\right)}^2=8+0+12=20$
${\left(\frac{\partial f}{\partial y}\right)}_{(1,0,2)}=7\left(1\right)=7$
${\left(\frac{\partial f}{\partial z}\right)}_{(1,0,2)}=6\left(1\right)\left(2\right)=12$

Substituting these values back into the gradient formula gives:
$\nabla f(1,0,2)=20\hat{i}+7\hat{j}+12\hat{k}$
This vector represents the direction of the maximum rate of increase of $f$ at the point $P$.