# Consider f(x,y) = y^2+x^2 y + xy . Find the critical points of the above function and...

## Question:

Consider {eq}f(x,y) = -y^2+x^2 y + xy {/eq}.

Find the critical points of the above function and evaluate them accordingly as 1) minimum; 2) maximum; or 3) saddle point.

## Maximum and Minimum of a Two Variables Function:

If there is a function of two variables, to find the relative extremes or saddle points it is necessary to apply the theorem of the first partial derivatives for critical points and the theorem of the second derivatives for maximums and minimums together with the criterion of the Hessian determinant.

## Answer and Explanation:

**Function.**

{eq}\displaystyle f (x,y) = y^2+x^2y+xy {/eq}

**Partial derivatives.**

{eq}\begin{array}{lll} \text{First derivatives.} &&& \text{Second derivatives.} \\ \displaystyle f _x = 2xy+y &&& \displaystyle f _{xx} = 2y \\ \displaystyle f _y = 2y+x^2+x &&& \displaystyle f _{yy} = 2 \\ &&& \displaystyle f _{xy} = 2x+1 \end{array} {/eq}

**Calculus of critical values.**

The critical values are obtained using the fact that

{eq}\displaystyle f _x=0 \\ \displaystyle f _y=0 {/eq}

We need to solve the following system of equations:

{eq}\displaystyle 2xy+y = 0 \\ \displaystyle 2y+x^2+x = 0 {/eq}

The solution to the previous system of equations is:

{eq}\displaystyle ( x_ 1 , y_ 1 ) = \left( 0 , 0 \right) \\ \displaystyle ( x_ 2 , y_ 2 ) = \left( -1 , 0 \right) \\ \displaystyle ( x_ 3 , y_ 3 ) = \left( {-\frac{1}{2}} , \frac{1}{8} \right) \\ {/eq}

The Hessian determinant is calculated using the formula

{eq}\displaystyle H(x,y) = f _{xx}(x_0,y_0) f _{yy}(x_0,y_0) - [ f _{xy}(x_0,y_0) ]^2 {/eq}

Now, we calculate the value of the second derivative and Hessian determinant at each critical point:

At {eq}\displaystyle ( x_ 1 , y_ 1 ) = \left( 0 , 0 \right) : {/eq}

{eq}\displaystyle f \left( 0 , 0 \right) = (0)^2+(0)^2(0)+(0)(0) = 0 \\ \displaystyle f _{xx} \left( 0 , 0 \right) = 2(0) = 0 \\ \displaystyle f _{yy} \left( 0 , 0 \right) = 2 = 2 \\ \displaystyle f _{xy} \left( 0 , 0 \right) = 2(0)+1 = 1 \\ \displaystyle H \left( 0 , 0 \right) = \left( 0 \right)\left( 2 \right)-\left( 1 \right)^2= -1 {/eq}

**Conclusion:**

Since {eq}\displaystyle f _{xx} \left( 0 , 0 \right) = 0 , H \left( 0 , 0 \right) < 0 , {/eq} the point {eq}\displaystyle ( x_ 1 , y_ 1 ) = \left( 0 , 0 \right) {/eq} conduits to saddle point .

At {eq}\displaystyle ( x_ 2 , y_ 2 ) = \left( -1 , 0 \right) : {/eq}

{eq}\displaystyle f \left( -1 , 0 \right) = (0)^2+(-1)^2(0)+(-1)(0) = 0 \\ \displaystyle f _{xx} \left( -1 , 0 \right) = 0 = 0 \\ \displaystyle f _{yy} \left( -1 , 0 \right) = 2 = 2 \\ \displaystyle f _{xy} \left( -1 , 0 \right) = -1 = -1 \\ \displaystyle H \left( -1 , 0 \right) = \left( 0 \right)\left( 2 \right)-\left( -1 \right)^2= -1 {/eq}

**Conclusion:**

Since {eq}\displaystyle f _{xx} ( -1 , 0 ) = 0 , H \left( -1 , 0 \right) = 0 , {/eq} the point {eq}\displaystyle ( x_ 2 , y_ 2 ) = \left( -1 , 0 \right) {/eq} conduits to saddle point .

At {eq}\displaystyle ( x_ 3 , y_ 3 ) = \left( {-\frac{1}{2}} , \frac{1}{8} \right) : {/eq}

{eq}\displaystyle f \left( {-\frac{1}{2}} , \frac{1}{8} \right) = \left(\frac{1}{8}\right)^2+\left(-\frac{1}{1}\right)^2\left(\frac{1}{8}\right)+\left(-\frac{1}{2}\right)\left(\frac{1}{8}\right) = {-\frac{1}{64}} \\ \displaystyle f _{xx} \left( {-\frac{1}{2}} , \frac{1}{8} \right) = 2\left(\frac{1}{8}\right) = \frac{1}{4} \\ \displaystyle f _{yy} \left( {-\frac{1}{2}} , \frac{1}{8} \right) = 2 = 2 \\ \displaystyle f _{xy} \left( {-\frac{1}{2}} , \frac{1}{8} \right) = 2\left(-\frac{1}{2}\right)+1 = 0 \\ \displaystyle H \left( {-\frac{1}{2}} , \frac{1}{8} \right) = \left( \frac{1}{4} \right)\left( 2 \right)-\left( 0 \right)^2= \frac{1}{2} {/eq}

**Conclusion:**

Since {eq}\displaystyle f _{xx} ( {-\frac{1}{2}} , \frac{1}{8} ) > 0 , H \left( {-\frac{1}{2}} , \frac{1}{8} \right) > 0 , {/eq} the point {eq}\displaystyle ( x_ 3 , y_ 3 ) = \left( {-\frac{1}{2}} , \frac{1}{8} \right) {/eq} conduits to minimum relative .