find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point.
f(x,y) = (1/2)x^2 + 2y^2 - 8y + 4x
For this question, it is first necessary to find the partial derivatives with respect to x and y and set them equal to zero:
\(\frac{df}{dx}\) = x + 4 = 0 \(\frac{df}{dy}\) = 4y - 8 = 0
Solving for those, we get: x = -4 y = 2 So our critical point is (-4, 2)
We can now find the 2nd partial derivatives for both x and y:
The 2nd derivative of 4 + x is equal to 1 The 2nd derivative of 4y - 8 is equal to 4 The mixed derivative is found by using the partial derivative of x and using that to determine y, but that would be equal to 0
With this, we can multiply the 2nd derivatives together to get 4
Thus, our D value is greater than 0, as is our fxx(x,y); so, we can say that f(x,y) has a relative minimum at (-4,2)