Question

Given the following equations, find the extremums and if they are local minima, maxima, or saddle points? [15 pts] a. \( f(x)=x^{3}-x^{2} \) b. \( z(x, y)=-x^{3}+2 x^{2}+3 y^{2}+2 y \)

Answer 1

For f(x)=
`x^(3)`-
`x^(2)` , the extremums are at x-0, where it has a local minimum. For z(x, y)=-
`x^(3)`+2
`x^(2)`+3
`y^(2)`+2 y , there are critical points at (0,0) and (2,-1). (0,0) is a saddle point, while (2,-1) represents a local maximum.

Explanation:

For f(x)=
`x^(3)`-
`x^(2)`, the critical point occurs at x = 0 after finding the derivative f'(x) = 3
`x^2` - 2x. Analyzing the behavior around this point through the second derivative f''(x) = 6x - 2 confirms it as a local minimum because f''(0) = -2 , indicating a concave-down shape.

Regarding z(x, y)=-
`x^(3)`+2
`x^(2)`+3
`y^(2)`+2 y, the partial derivatives
`\( (\partial z)/(\partial x)` = -3
`x^2` + 4x and
`\( (\partial z)/(\partial y)` = 6y + 2 help find critical points. Setting both partial derivatives to zero gives x = 0 and y = -1/3 or (0, -1/3) as critical points.

Evaluating the second derivatives,
`\( (\partial^2 z)/(\partial x^2)` = -6x + 4 and
`\( (\partial^2 z)/(\partial y^2)` = 6 , we check the points' nature. At (0,0), the second derivatives show it as a saddle point due to
`\( (\partial^2 z)/(\partial x^2)` (0,0) = 4 and
`\( (\partial^2 z)/(\partial y^2)`(0,0) = 6. However, at (2,-1),
`\( (\partial^2 z)/(\partial x^2)`(2,-1) = -8 and
`\( (\partial^2 z)/(\partial y^2)`(2,-1) = 6 , confirming it as a local maximum due to a concave-down shape in the x-direction and a concave-up shape in the y-direction.