INSTRUCTOR: Find the direction for which the directional derivative of g of x comma y equal to 4x minus xy plus 2y squared at the point negative 2 comma 3 is a maximum. What is the maximum value?

The directional derivative is going to be maximized in the same direction as the gradient of that function at that point. So we're going to start by finding the gradient. So the gradient of g of x comma y is going to be equal to g sub x i plus g sub y j.

Well, g sub x is 4 minus y. This will be equal to parentheses 4 minus y, close parentheses, i plus, open parentheses, negative x plus 4y, which is g sub y, close parentheses, j.

And evaluating this at the point negative 2, 3, the gradient of g, evaluated at negative 2, 3 is going to be equal to i plus 14j.

Now, we can find the magnitude of that vector to give us the maximum value. This maximum value is equal to the square root of 1 squared plus 14 squared. And that is equal to the square root of 197. There's our maximum value.

But to find the direction, we need to find a unit vector that's in the same direction as the gradient at that point. Now, we already have the magnitudes. We're going to hold on to that. So the magnitude of the gradient of g evaluated at negative 2, 3 equal to the square root of 197.

So our unit vector-- I'm going to go ahead and just call it u-- is going to be 1 over the square root of 197, as our i component, plus 14 over the square root of 197 as our j component. Taking the vector, or the gradient, there, dividing it by its magnitude. And because this is in ij format, this tells us the cosine and the sine of our vector.

I'm going to use the fact that, in this case, the sine of theta is equal to 14 over the square root of 197. And if we take the arcsine of this, of both sides, theta is approximately 1.499 radians. And that is the direction of r maximum with a maximum value of square root of 197.