1.
Find the absolute maximum and minimum values of
\begin{equation*}
f(x,y)=x^2-xy+y^2-3y
\end{equation*}
on the region bounded by the \(x\)- and \(y\)-axes and the line \(x+y=4\text{.}\)
Solution.
The gradient is \(\nabla f = \langle 2x - y,\; -x + 2y - 3\rangle\text{,}\) which vanishes only at \((1,2)\text{,}\) a point in the interior of the region. There \(f(1,2) = -3\text{.}\) Since this is the only critical point on all of \(\mathbb{R}^2\) and it is a minimum (by the second derivative test), it is a global minimum. We now examine the three boundary segments to look for possible maxima.
On the \(x\)-axis with \(0 \le x \le 4\text{,}\) we have \(f(x,0) = x^2\text{,}\) which is maximized at \(f(4,0) = 16\text{.}\)
On the \(y\)-axis with \(0 \le y \le 4\text{,}\) we have \(f(0,y) = y^2 - 3y\text{,}\) which is maximized at \(f(0,4) = 4\text{.}\)
Finally, on the line \(x = 4 - y\) with \(0 \le y \le 4\text{,}\) we have \(f(4-y,\,y) = 3y^2 - 15y + 16\text{,}\) which is maximized at \(f(4,0) = 16\text{.}\)
Altogether, there is an absolute minimum \(f(1,2) = -3\) and an absolute maximum \(f(4,0) = 16\text{.}\)
