Find the maximum and minimum values of the function fxy2x23y

Find the maximum and minimum values of the function f(x,y)=2x2+3y24x5 on the domain x2+y2400. As usual, ignore unneeded answer blanks, and list points lexicographically. Maximum value is , occuring at ( , ), and ( , ). Minimum value is , occuring at ( , ) and ( , ).

Solution

given f(x,y)=2x^2+3y^24x5 on the domain x^2+y^2400.

now we have to find fx , fy

fx= 4x-4

fx=0

4x-4 =0

so x=1

fy=6y=0 so(1,0) is a critical point and f(1,0)=-7
now you have to take into account the points of x^2+y^2=400

y^2= 400-x^2so f(x) = 2x^2+3(400 - x^2)-4x-5

f(x) = 2x^2+1200 -3x^2 -4x- 5

f(x) = -x^2 -4x +1195

for f(x) maximum we have to find out f\'(x)

f\'(x)= -2x -4 =0

so x=-2

f(x) has a maximum at x=-2 and x^2+y^2= 400 so

y^2 = 400-4

y^2=396
f(x,y) = 8+3(396)+8-5

f(x,y) = 8+1188+3 =1199 maximum

maximum vale is 1199 is occuring at (-2,6sqrt(11) and (-2 -6sqrt(11) )

minimum value is -7 occuring at (1,0)