Find the maximum rate of change

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SUMMARY

The maximum rate of change of the function f(x,y) = (3 y^5)/x at the point (1,2) is determined using the gradient. The correct gradient is calculated as ∇f(x,y) = <-3y^5*x^-2, 15y^4*x^-1>. Evaluating this gradient at the point (1,2) yields the vector <-96, 240>. The maximum rate of change is then found by taking the dot product of the gradient with the unit vector in the direction of interest, resulting in a value of 171.73.

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andyk23
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Find the maximum rate of change of the function f(x,y) = (3 y^5)/x at the point (1,2)

First I took the gradient of f(x,y)=<-3y^5*x^-2,15y^4*x^-1>
and took the pt <1/sqrt(5),2/sqrt(5)>
Then <-3y^5*x^-2,15y^4*x^-1>*<1/sqrt(5),2/sqrt(5)>
the answer i get is 171.73
I'm not sure where my error is
 
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andyk23 said:
Find the maximum rate of change of the function f(x,y) = (3 y^5)/x at the point (1,2)

First I took the gradient of f(x,y)=<-3y^5*x^-2,15y^4*x^-1>
and took the pt <1/sqrt(5),2/sqrt(5)>
Then <-3y^5*x^-2,15y^4*x^-1>*<1/sqrt(5),2/sqrt(5)>
the answer i get is 171.73
I'm not sure where my error is
Why not evaluate the gradient at the point (1, 2) ?
 
Sorry I evaluated at the gradient(1,2)=<-96,240> and then <-96,240>*<1/sqrt(5),2/sqrt(5)>
 

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