Find the maximum rate of change off at the given point and the direction in which it occurs. f(x, y) = 4y √x, (4, 1)

Maximum rate change |f| = √65

Direction = ( 1, 8 ) / √65

i.e Maximum rate change is √65 and it occurs in a direction of ( 1, 8 )

Step-by-step explanation:

Given that;

f(x, y) = 4y √x, (4, 1)

{ x=4 and y=1 }

Maximum rate of change occurs with the gradient vector;

f = [ df/dx, df/dy ] = [ 2y/√x, 4√x ]

we substitute in our value of x and y

f =  2(1)/√4, 4√4

f = 2/√4, 4√4

f = ( 1, 8 )

Maximum rate change |f| = | ( 1, 8 ) | = √( 1² + 8² ) = √(1 + 64)

Maximum rate change |f| = √65

Direction = f / |f|

we substitute

Direction = ( 1, 8 ) / √65

i.e Maximum rate change is √65 and it occurs in a direction of ( 1, 8 )

constants mean the stopped points or the points on a graph that nothing is changing, hope this

step-by-step explanation:

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