# Find the critical point of the function: f(x,y)=x^2+y^2-xy+1.5x. Use the second partial...

## Question:

Find the critical point of the function: {eq}f(x,y)=x^2+y^2-xy+1.5x{/eq}

Use the second partial derivative test to identify the critical point.

## Critical Points of Function:

The critical points of a function of two real variables {eq}f(x,y) {/eq}

are found setting the first partial derivatives to zero

{eq}f_x(x,y) = 0 \\ f_y(x,y) = 0 {/eq}

If {eq}(x_0,y_0) {/eq} is a critical point, we compute the determinant of Hessian matrix

{eq}D = f_{xx}(x_0,y_0) f_{yy}(x_0,y_0) -f_{xy}^2(x_0,y_0) {/eq}

Then, we classifiy the critical point as

{eq}\displaystyle D > 0 \;, f_{xx}(x_0,y_0) > 0 \rightarrow \; minimum \\ \displaystyle D > 0 \;, f_{xx}(x_0,y_0) < 0 \rightarrow \; maximum \\ \displaystyle D < 0 \rightarrow \; saddle \; point \\ \displaystyle D=0 \; no \; conclusion {/eq},

The critical points of the function

{eq}f(x,y) =x^2+y^2-xy+1.5x {/eq}

are found setting its first partial derivatives to zero

{eq}\displaystyle f_x(x,y) =2x - y + 1.5 = 0 \\ \displaystyle f_y(x,y) = 2y -x = 0 {/eq}

Upon solving the linear system by substitution

{eq}2y -x = 0 \Rightarrow x = 2y \\ 2x - y + 1.5 = 0 \Rightarrow 4y - y +1.5 = 0 \\ \Rightarrow y=-0.5 \Rightarrow x=-1 {/eq}

we get the critical point {eq}(-1,-0.5) {/eq}

By computing the determinant of hessian matrix we find

{eq}\displaystyle f_{xx}=2 \\ \displaystyle f_{yy}=2 \\ \displaystyle f_{xy}= -1 \\ \displaystyle D(-1,-0.5) = f_{xx}(-1,-0.5) f_{yy}(-1,-0.5) -f_{xy}^2(-1,-0.5) = 2(2)- (-1)^2 =3. {/eq}

Since the determinant is positive and {eq}f_{xx}=2 > 0 {/eq} the point {eq}(-1,-0.5) {/eq} is a local minimum.