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.

This question is based on the absolute maximum and absolute minimum.

We get this by differentiating the terms.

Given:

f(x,y) =   , bounded by the lines x=0,y=2,y=2x in the first quadrant,bounded by the lines x=0,y=2,y=2x in the first quadrant.

We need to determined the absolute maximum and absolute minimum of the function.

Now, partial differentiating wrt x and y.

Now, 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).

Now, find f(0, 0) = 1 (MAX) and f(0, 2) = -3.

Now check on the boundary lines themselves.

If x = 0, then we get,

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

If y = 2, then we get,

f(x,2) = with a maximum of -5 when x = 1.

If y = 2x, then we get,

f(x,2x) = with the same maximum of -5 when     x = 1.

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

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

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

If y = 2, then

with a maximum of -5 when x = 1.

If y = 2x, then

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

no pictures of the graph