Question

Use a graph and/or level curves to estimate the local maximum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisely.

\(f(x,y) = 3{x^2}y + {y^3} - 3{x^2} - 3{y^2} + 2\)

Short answer

Local maximum point is at\((0,0)\),local minimum\((0,2)\)& saddle point is at\((0,0)\). Saddle points are\((1,1),( - 1,1)\)

Step-by-step solution

Second derivative test

Suppose the second partial derivatives of\(f\)are continuous on a disk with center\({\bf{(a,b)}}\), and suppose that\({f_x}(a,b) = 0\)and\({f_y}(a,b) = 0\). Let

\(D = D(a,b) = {f_{xx}}(a,b){f_{yy}}(a,b) - {\left( {{f_{xy}}(a,b)} \right)^2}\)

(a) If\(D > 0\)and\({f_{xx}}(a,b) > 0\), then\(f(a,b)\)is a local minimum.

(b) If\(D > 0\)and\({f_{xx}}(a,b) < 0\), then\(f(a,b)\)is a local maximum.

(c) If\(D < 0\), then\(f(a,b)\)is not a local maximum or minimum.

Using graph

Consider the following function\(f(x,y) = 3{x^2}y + {y^3} - 3{x^2} - 3{y^2} + 2\).

Using graphing calculator the graph is:

By graph it seems that\(f\)has a local maximum\(f(0,0) \approx 2\), and a local minimum\(f(0,2) \approx - 2\)and saddle points near\(( \pm 1,1).\)

Find partial derivative

Use the Calculus to find the precise values.

We have\(f(x,y) = 3{x^2}y + {y^3} - 3{x^2} - 3{y^2} + 2\)

The partial derivative of \(f(x,y)\)with respect to\(x\,\& \,y\)is given by:

\(\begin{aligned}{l}{f_x}(x,y) = 6xy - 6x\\{f_y}(x,y) = 3{y^2} - 6y\end{aligned}\)

Find critical point

Put\({f_x} = 0,{f_y} = 0\)to find critical points.

\(\begin{aligned}{c}{f_x}(x,y) = 0\\6xy - 6x = 0\\x = 0{\rm{ or }}y = 1\end{aligned}\)

And

\(\begin{aligned}{c}{f_y} = 0\\3{y^2} - 6y = 0\\y(3y - 6) = 0\\y = 0{\rm{ or }}y = 2\end{aligned}\)

Plugging\(y = 1\)into the equation and equating this expression to zero,

\(\begin{aligned}{c}3{x^2} + 3{(1)^2} - 6(1) = 0\\3{x^2} = 3\\x = \pm 1\end{aligned}\)

Thus, the four critical points of the function are \((0,0),(0,2),(1,1)\)and\(( - 1,1)\).

Find second order derivative

For\(f(x,y) = 3{x^2}y + {y^3} - 3{x^2} - 3{y^2} + 2\)

Differential two times with respect to\(x\),

\(\begin{aligned}{c}{f_x}(x,y) = 6xy - 6x\\{f_{xx}}(x,y) = 6y - 6\\{f_{xy}}(x,y) = 6x\\{f_{yy}}(x,y) = 6y - 6\end{aligned}\)

Calculate discriminant

The value of \(D\)is,

\(\begin{aligned}{c}D(a,b) = {f_{xx}}(a,b){f_{yy}}(a,b) - {\left( {{f_{xy}}(a,b)} \right)^2}\\D = (6y - 6)(6y - 6) - {(6x)^2}\\ = 36{(y - 1)^2} - 36{x^2}\\ = 36\left( {{{(y - 1)}^2} - {x^2}} \right)\end{aligned}\)

Evaluate discriminant at critical point

At the critical point\((0,0)\),

\(\begin{aligned}{c}D = 36\left( {{{(0 - 1)}^2} - {0^2}} \right)\\ = 36\end{aligned}\)

And

\(\begin{aligned}{c}{f_{xx}}(x,y) = 6y - 6\\{f_{xx}} = 6(0) - 6\\ = - 6\end{aligned}\)

Since\(D > 0\) and\({f_{xx}} < 0\), therefore the point\((0,0)\) is a local maximum.

\(\begin{aligned}{c}f(0,0) = 3{(0)^2}(0) + {(0)^3} - 3{(0)^2} - 3{(0)^2} + 2\\ = 2\end{aligned}\)

Evaluate discriminant at critical point

At the critical point\((0,2)\),

\(\begin{aligned}{c}D(0,2) = 36\left( {{{(2 - 1)}^2} - {0^2}} \right)\\ = 36\end{aligned}\)

And

\(\begin{aligned}{c}{f_{xx}}(x,y) = 6y - 6\\ = 6(2) - 6\\ = 6\end{aligned}\)

Since \(D > 0\)and\({f_{xx}} > 0\), therefore the point \((0,2)\)is a local minimum.

\(\begin{aligned}{c}f(0,2) = 3{(0)^2}(2) + {(2)^3} - 3{(0)^2} - 3{(2)^2} + 2\\ = 8 - 12 + 2\\ = - 2\end{aligned}\)

Evaluate discriminant at critical point

At the critical point\((1,1)\),

\(\begin{aligned}{c}D(1,1) = 36\left( {{{(1 - 1)}^2} - {1^2}} \right)\\ = - 36\end{aligned}\)

Since\(D < 0\), therefore the point \((1,1)\)is a saddle point.

At the critical point\(( - 1,1)\),

\(\begin{aligned}{c}D = 36\left( {{{(1 - 1)}^2} - {{( - 1)}^2}} \right)\\ = - 36\end{aligned}\)

Since\(D < 0\), therefore the point \(( - 1,1)\)is a saddle point.