How do you find the critical point and determine whether it is a local maximum, local minimum, or neither for $f(x,y)={x}^{2}+4x+{y}^{2}$ ?

Addison Gross
2022-01-25
Answered

How do you find the critical point and determine whether it is a local maximum, local minimum, or neither for $f(x,y)={x}^{2}+4x+{y}^{2}$ ?

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trasahed

Answered 2022-01-26
Author has **14** answers

The first-order partial derivatives of $z=f(x,y)={x}^{2}+4x+{y}^{2}$ are $\frac{\partial z}{\partial x}=2x+4$ and $\frac{\partial z}{\partial y}=2y$ .

Setting both of these equal to zero results in a system of equations whose unique solution is clearly$(x,y)=(-2,0)$ , so this is the unique critical point of $f$ .

The second-order partials are$\frac{{\partial}^{2}z}{\partial {x}^{2}}=2,\frac{{\partial}^{2}z}{\partial {y}^{2}}=2$ and $\frac{{\partial}^{2}z}{\partial x\partial y}=\frac{{\partial}^{2}z}{\partial y\partial x}=0$

This makes the discriminant for the (multivariable) Second Derivative Test equal to

$D=\frac{{\partial}^{2}z}{\partial {x}^{2}}\cdot \frac{{\partial}^{2}z}{\partial {y}^{2}}-{\left(\frac{{\partial}^{2}z}{\partial x\partial y}\right)}^{2}=2\cdot 2-{0}^{2}=4>0$

which means the critical point is either a local max or a local min (it's not a saddle point).

Since$\frac{{\partial}^{2}z}{\partial {x}^{2}}=2>0$ , the critical point is a local min.

Setting both of these equal to zero results in a system of equations whose unique solution is clearly

The second-order partials are

This makes the discriminant for the (multivariable) Second Derivative Test equal to

which means the critical point is either a local max or a local min (it's not a saddle point).

Since

lilwhitelieyc

Answered 2022-01-27
Author has **10** answers

The critical point makes both partial derivatives 0 (simultaneously).

For this function there is one critical point:

To determine whether f has a local minimum, maximum or neither at this point we apply the second derivative test for functions of two variables. (Well, we try to apply it. It does not always give an answer.)

Evaluate the second partials at the critical point (In this case they are all constant, but in general we cannot skip this step.)

At the critical point

Calculate

Apply the second derivative test:

Since D is positive, we look at A and with

f

f has a local minimum of−4 at

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