Answer:
(-24, -8)
Step-by-step explanation:
Let us recall that when we have a function f
[tex]\large f:\mathbb{R}^2\rightarrow \mathbb{R}\\f(x,y)=z[/tex]
if the gradient of f at a given point (x,y) exists, then the gradient of f at this point (x,y) gives the direction of maximum rate of increasing and minus the gradient of f at this point gives the direction of maximum rate of decreasing. That is
[tex]\large \nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})[/tex]
at the point (x,y) gives the direction of maximum rate of increasing
[tex]\large -\nabla f[/tex]
at the point (x,y) gives the direction of maximum rate of decreasing
In this case we have
[tex]\large f(x,y)=100-4x^2-2y^2[/tex]
and we want to find the direction of fastest speed of decreasing at the point (-3,-2)
[tex]\large \nabla f(x,y)=(-8x,-4y) \Rightarrow -\nabla f=(8x,4y)[/tex]
at the point (-3,-2) minus the gradient equals
[tex]\large -\nabla f(-3,-2)=(-24,-8)[/tex]
hence the vector (-24,-8) points in the direction with the greatest rate of decreasing, and they should start their descent in that direction.
