Finding Absolute Extrema

Find the absolute maximum and the absolute minimum of \[ \begin{equation*} z=f(x,y)=2x-2xy+y^{2} \end{equation*} \]

whose domain is the region defined by \(0\leq x\leq 4\) and \(0\leq y\leq 3\).

Solution  The function \(f\) is continuous on its domain, which is a closed, bounded set. Then the Extreme Value Theorem guarantees that \(f\) has an absolute maximum and an absolute minimum on its domain. To find them, we first find the critical points of \(f,\) namely the solutions of the system of equations \[ \left\{\begin{array}{l} f_{x}(x,y)=2-2y=0\\ f_{y}(x,y)=-2x+2y=0 \end{array}\right. \]

Figure 18 \(z=f(x,y)=2x-2xy+y^{2},\) \(0\leq x\leq 4,\) \(0\leq y\leq 3\)

Solving the system of equations, we find that the only critical point is \((1,1)\). The value of \(f\) at \((1,1)\) is \[ f(1,1)=2(1) -2(1) (1) +1^{2}=1 \]

The domain of \(f\) is the set \(0\leq x \leq 4\), \(0\leq y\leq 3\). The boundary of the domain of \(f\) consists of the four line segments \(L_1\), \(L_2\), \(L_3\), and \(L_4\) shown in Figure 18. We evaluate \(f\) on each line segment.

  • On \(\boldsymbol L_{\bf 1}\): \(x\) is in the interval \(\left[ 0,4\right]\) and \(y=0\). The function \(f(x,y)=f(x,0)=2x\) is increasing on \(0\leq x\leq 4\), so its extreme values occur at the endpoints \(0\) and \(4.\) \[ \hbox{At }x=0, f(0,0)=0\qquad \hbox{and} \qquad \hbox{at}\ x=4, f(4,0)=8 \]

  • On \(\boldsymbol L_{\bf 2}\): \(x=4\) and \(y\) is in the interval \([0,3] \). The function \(f(x,y)=f(4,y)= 8-8y+y^{2}\). To find the extreme values of \(f\) on \( [0,3] \), we begin by finding the critical number(s) of the function \(g(y) =8-8y+y^{2}.\) That is, we find where \(g' (y) =0.\) \[ \begin{eqnarray*} g' (y) &=&-8+2y=0 \\[3pt] y &=&4 \end{eqnarray*} \]

    Since \(4\) is not in the interval \([0,3] ,\) the extreme values occur at the endpoints 0 and 3. \[ \hbox{At }y=0\hbox{, }\ f(4,0)=8\qquad \hbox{and} \qquad \hbox{at}y=3,\hbox{ }f(4,3)=-7 \]

  • On \(\boldsymbol L_{\bf 3}\): \( x\) is in the interval \([ 0,4] \) and \(y=3\). The function \(f(x,y)=f(x,3)= 2x-6x+9=-4x+9\) is decreasing on \(0\leq x\leq 4\), so its extreme values occur at the endpoints \(0\) and \(4.\) \[ \hbox{At }x=0, f(0,3)=9\qquad \hbox{and} \qquad \hbox{at }x=4, f(4,3)=-7 \]

  • On \(\boldsymbol L_{\bf 4}\): \(x=0\) and \(y\) is in the interval \([0,3] .\) The function \(f(x,y)=f(0,y)=y^{2}\) is increasing on \(0\leq y\leq 3\), so its extreme values occur at the endpoints \(0\) and \(3.\) \[ \hbox{At }y=0, f(0,0)=0\qquad\hbox {and}\qquad \hbox{at }y=3,\hbox{ }f(0,3)=9 \]

    •

    885

    The values of \(f\) at the critical point \((1,1)\) and at the extreme values on the boundary are

    Point \((1,1)\) \(( 0,0)\) \(( 4,0)\) \(( 4,3)\) \(( 0,3)\)
    Value \(1\) \(0\) \(8\) \(-7\) \(9\)

    The absolute maximum value of \(f\) is \(f(0,3)=9\); the absolute minimum value is \(f(4,3)=-7\).