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Vera_Pavlovna [14]
3 years ago
15

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 t

he lines x=0,y=2,y=2xin the first quadrant.
Mathematics
1 answer:
marusya05 [52]3 years ago
6 0

First check for the critical points of <em>f</em> 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 <em>y</em> = 2<em>x</em> ; at this point, we get a value of <em>f</em>(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 <em>f</em>(0, 0) = 1 (MAX) and <em>f</em>(0, 2) = -3.

Now check on the boundary lines themselves. If <em>x</em> = 0, then

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

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

If <em>y</em> = 2, then

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

with a maximum of -5 when <em>x</em> = 1.

If <em>y</em> = 2<em>x</em>, then

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

with the same maximum of -5 when <em>x</em> = 1.

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