Question

Find the absolute maximum and absolute minimum of the function f(x,y)=2x2−4x+y2−4y+1 on the closed triangular plate bounded by the lines x=0,y=2,y=2xin the first quadrant.

Answer 1

First check for the critical points of f by checking where the first-order derivatives vanish.

$\dfrac{\partial f}{\partial x}=4x-4=0\implies x=1$

$\dfrac{\partial f}{\partial y}=2y-4=0\implies y=2$

Notice how the point (1, 2) lies on the line y = 2x ; at this point, we get a value of f(1, 2) = -5 (MIN).

Next, check the points where the boundary lines intersect, which occurs at the points (0, 0), (0, 2), and (1, 2). We already checked the last one. We find f(0, 0) = 1 (MAX) and f(0, 2) = -3.

Now check on the boundary lines themselves. If x = 0, then

$f(0,y)=y^2-4y+1=(y-2)^2-3$

which has a maximum value of -3 when y = 2 (so we get the same critical point as before).

If y = 2, then

$f(x, 2)=2x^2-4x-3=2(x-1)^2-5$

with a maximum of -5 when x = 1.

If y = 2x, then

$f(x,2x)=6x^2-12x+1=6(x-1)^2-5$

with the same maximum of -5 when x = 1.