Question

NEED HELP! I have just 30 minutes to solve on my online calculus 2 exam, PLS HELP QUICKLY.

Find the absolute maxima and minima of the function on the given domain. f(x, y) = x2 + 10x + y2 + 20y +5 on the rectangular region-1sxs1, -2sys2 Absolute maximum: 60 at (1, 2); absolute minimum: 5 at (0,0) Absolute maximum: 73 at (2, 2); absolute minimum: 5 at (0,0) Absolute maximum: 60 at (1, 2); absolute minimum: -40 at (-1, -2) Absolute maximum: 73 at (2.2); absolute minimum: -40 at (-1, -2)

Answer 1

feel free to ask any doubt please. If you don't have any doubt please like.

11-1 e (72) +20 cm (4) + 5) (42) + 10,00 (ca) + Answer – given f(x, y) = x2 + 10x + y2 + 20y + 5 given region is -1 < x < 1,-2 <y < 2 # Suppose that (a,b) is a critical point of f(x, y) and that the second order partial derivatives are continuous in some region that contains (a,b) Next define, D = D(a,b) = fxx(a,b)fyy(a,b) – [fxy(a, b)]? We then have the following classifications of the critical point. (1) (1) If D > 0 and fxx(a,b) > 0 then there is a relative minimum at (a,b) (2) If D > 0 and fxx(a,b) < 0 then there is a relative maximum at (a,b) (3)If D <0 then the point (a,b) is a saddle point. d Basic Rules -(1) --" = n. t dt d d (2) -c f(t) = 0 f(t),c = constant dt dt a д a Now, fx(x, y) - (x2) + 10—(x) + 2(72) + 20—(y) + дх дх дх дах дх = 2x + 10(1) +0+ 0 = 2x + 10 = fx(x, y) = 2x + 10 a a = fxx(x, y) = = 2(x) + -(10) = 2(1) + 0 = 2 дх a a а a a similarly, fy(x, y) = (x2) + 10(x) + (72) +20=(y) + (5) = 0+0+ 2y + 20(1) + 0 = 2y + 20 fy(x, y) = 2y + 20 a a & fyy(x, y) = 2(y) + -(20) = 2(1) + 0 = 2 a д & f xy(x, y) 2-(x) + (10) = 0 + 0 = 0 ду Now, for critical points we must have fx(x, y) = 0, fy(x, y) = 0 so, 2x + 10 = 0 3 x = -5 & 2y + 20 = 0 3y = -10 so point is (-5, -10) but it does not lie in the given domain because – 5 ¢[-1,1] & - 10 € (-2,2] let us find D i.e. D = (2)(2) – 0) = 4 >0 & f xx(x, y) > 0 for all points. Therefore we just need to know the value of the functions at that point. so, let us take corner points & origin i.e.(-1,-2), (-1,2), (1, -2), (1,2)& (0,0) Now(a)at (-1,-2), f(-1, -2) = (-1)2 + 10(-1) + (-2)2 + 20(-2) + 5 = – 40 (b)at (-1,2), f(-1,22) = (-1)2 + 10(-1) + (2)2 + 20(2) + 5 = 40 (c)at (1, 2), f(1, 2) = (1)2 + 10(1) + (2)2 + 20(2) + 5 = 60 (d)at (1, -2), f(1, -2) = (1)2 + 10(1) + (-2)2 + 20(-2) + 5 = – 20 (e)at (0,0), f(0,0) = 0 +0+0+0+5 = 5 so absolute max. = 60 at (1,2) absolute min -40 at (-1,-2) #Ans..