Homework Help: Why this maximization approach fails?

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1. Oct 14, 2015

friendbobbiny

1. The problem statement, all variables and given/known data
Find all points at which the direction of fastest change of the function $$f(x,y) = x^2 + y^2 -2x - 2y$$is in the direction of <1,1>.

2. Relevant equations
$$<\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}>$$

3. The attempt at a solution
$$\frac{\nabla f}{|\nabla f|}$$ = <1,1>

This doesn't work but

$$\frac{\delta f}{\delta x} = \frac{\delta f}{\delta y}$$

does. This latter approach makes sense. The partial derivatives at x and y have the same value. The former approach should work, however. In general, $$\nabla f$$ gives the direction of maximum change.

2. Oct 14, 2015

friendbobbiny

NVM

set $$\frac{\nabla f}{|\nabla f|}$$ to <root(2)/2, root(2)/2>

Resolved!

3. Oct 14, 2015

Staff: Mentor

Your function is one with two variables, so the gradient should be a vector with two components, not three.
$$\nabla f = <\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}>$$
Do you have a feel for what the surface for this function looks like? The graph of the surface is a paraboloid that opens upward.

4. Oct 15, 2015

HallsofIvy

Just a note on Latex. First, your $\nabla F$ belongs outside the < > denoting the vector. Second you can use "\partial" to indicate partial derivatives rather than "\delta":
$$\nabla f = <\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z}>$$