Absolute Minimums and Maximums — Question 3

Question 3 – Solution

Strategy. The compact triangle guarantees extrema; compare the interior stationary value with the three edge restrictions.

Step 1: Interior $$f_x=y(4-2x-y),\qquad f_y=x(4-x-2y).$$ In the interior $x,y>0$, so $$2x+y=4,\qquad x+2y=4.$$ Thus $x=y=4/3$, and $$f\left(\frac 43,\frac 43\right) =\frac 43\frac 43\frac 43=\boxed{\frac{64}{27}}.$$

Step 2: Boundary On $x=0$ or $y=0$, one factor vanishes. On $x+y=4$, the third factor vanishes. Hence $$f=0\quad\text{on all three edges, including the vertices}.$$

Step 3: Comparison Inside $T$, all three factors $x$, $y$, and $4-x-y$ are positive, so $f>0$. Therefore $$\boxed{f_{\min}=0\text{ at every boundary point of }T},$$ $$\boxed{f_{\max}=\frac{64}{27}\text{ at }(4/3,4/3)}.$$ The candidate list is complete because the polynomial is continuous and $T$ is closed and bounded.