Question

What is the absolute minimum value of f(x,y) on the region
D: 0=<x=<4 and
0=<y=<2

If f(x,y)=x^2-2xy+4y^2-4x-2y+24 ?

Answer 1

o what is the absolute minimum Calue of ffxy) Di OSX54 and Os 42 on the region If f(x,y) = x-axythy 4x-24+24. solution :- Given the function facy): 220cy +44–42-24+24. and given magic on D: osx<4 and ocy2 first cse find critical parts fx and fy are make it zero 23 - 28-4-0 ie fx = Sy -27+84-2-0 from @t Gy-6=0 y=1 then apply in@ x=3 =0 Y=2- (4,2) 18 The critical pontein) = Bil is in the regim (0:2) and it is in the intersected (OP) in the givern segion; 'then f(3,1)= 17. 24 40) Y-0

nad find absolute Extrema for g) of from Gwen region of hedende OS X 54 ; Osus2 can then be defined as follows Top y=2 Bottom y=0 03834 OS X su osysa osysa Right x=4 deft ISO - Dao we need to analyze each of sides to get absolite extrema for f(x,y) that might occur on the boundars 0 Top 4=2, Osas4 then g(x) = f(0,2)=3"-42 +16-42-4+24 go) = -88 +36 22-8 20 now find absolute extrema for g(x) as from using Criti Cel points ie f(a)=0 ie 1:4 critical point is (4,2) then $(4,2)= 20. Botlom 4-0 iosas4 then 9000 = $(2,0) = x -42 +24

using Critical points ie gazo ie 21-4 = 0 2:2 the critical point is (2,0) 11 - :. Cy) = f(2,0) = 20 Right x=4, osys2 Then hy) = f(4,4) 16-84+47916-24 +24 hry) 44107+24 now find absolute Extrema for hly) as from using critical points ie licy) =0 87-10 = g=s Critical point then f(4,5) - ce -0 is (4, 4 Left a=0 ocy<2 then hly) = f(0,y)= 4y-24 +24 now find absolute Extrema for hry) as from using critical points ie Gcy) = 0 By-9=0 y= the critical point is (0,4) then f(0,4) = ce 95 9

critical -17 on the summary of all above points and Each of end points are (311), (4,2), (2,0), (4,5,60 the function lalues are f(3,1) f(4,2)=20 f(20) = 20 +(4:7 frott from the above values absolute minimum value is it at (3,1) for given function of f(x,y). 95