Question

Find the absolute maximum and minimum values of\(f\)the set\(D\).

\(f(x,y) = x{y^2};\,\,D = \left\{ {(x,y)\mid x \ge 0,y \ge 0,{x^2} + {y^2} \le 3} \right\}\)

Short answer

Absolute minimum is\(\;\;0{\rm{ }}.\)& absolute maximum is\(2\).

Step-by-step solution

Definition

The highest point of a function within the entire domain is known as the absolute maxima of the function whereas the lowest point of the function within the entire domain of the function, is known as the absolute minima of the function.

Region

The region is shown below:

Find partial derivative

Consider the following function\(f(x,y) = x{y^2}\;\;\; \cdots \cdots (1)\)

And the domain is\(D = \left\{ {(x,y)\mid x \ge 0,y \ge 0,{x^2} + {y^2} \le 3} \right\}.\)

Differentiate (1) with respect to\(x\,\& \,y\).

\(\begin{aligned}{l}{f_x} = {y^2}\\{f_y} = 2xy\end{aligned}\)

Find critical point

For critical point put\({f_x} = 0,{f_y} = 0\)

\(\begin{aligned}{*{20}{r}}{{y^2} = 0}\\{y = 0}\end{aligned}\)

And:

\(\begin{aligned}{l}2xy = 0\\x = 0{\rm{ or }}y = 0\end{aligned}\)

Therefore, the only critical point of the function \(f(x,y) = x{y^2}\)is\((0,0)\).

Substitute the critical point \((0,0)\)in\(f(x,y) = x{y^2}\), to get value of function at critical point.

\(f(0,0) = 0\)

Evaluate function on boundary

The boundary of the region is\({x^2} + {y^2} = 3\).

\(\begin{aligned}{c}{x^2} + {y^2} = 3\\{y^2} = 3 - {x^2},0 \le x \le \sqrt 3 \end{aligned}\)

Substitute in equation (1) to get,

\(\begin{aligned}{c}f(x,y) = x{y^2}\\f(x,x) = x\left( {3 - {x^2}} \right)\\ = 3x - {x^3}\end{aligned}\)

Let \(f(x,x) = g(x) = 3x - {x^3}\)

Find partial derivative

The partial derivative of\(g\)with respect to\(x\,\& \,y\)is,

\(\begin{aligned}{c}{g_x}(x) = \frac{\partial }{{\partial x}}\left( {3x - {x^3}} \right)\\ = 3 - 3{x^2}\end{aligned}\)

And

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

But, the value of \(x\)ranges from\(\;0\)to\(\sqrt 3 \).

So, the value\(x = - 1\)will be omitted So, \(x = 1\) is accepted.

Find critical points

When\(x = 1\)find\(y\):

\(\begin{aligned}{c}{y^2} = 3 - {x^2}\\{y^2} = 3 - {(1)^2}\\y = \pm \sqrt 2 \end{aligned}\)

The critical points are\((1, - \sqrt 2 )\)and\((1,\sqrt 2 )\)

Evaluate function on critical points

Evaluate the values of\(f\) on critical point.

\(\begin{aligned}{c}f(1, - \sqrt 2 ) = (1){( - \sqrt 2 )^2}\\ = 2\\f(1,\sqrt 2 ) = (1){(\sqrt 2 )^2}\\ = 2\end{aligned}\)

Evaluate function on boundary

The boundary points of\(D\)are\((0,0),(0,\sqrt 3 ),(\sqrt 3 ,0)\).

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

So, the absolute maximum is\(2\)and absolute minimum is\(\;\;0{\rm{ }}.\)