Find the minimum value of f(x, y, z) = 2x2 + 3y2 + 2z2 when the domain is constrained to the planes 2x + y + 4z = -5 and 5x + 3y - 3z = -4 You may round to the nearest hundredth if necessary.Find the volume of the solid bounded by the paraboloids z=-5+4+2z+2yand z =9 3z 3y. Round your answer to four decimal places. \f4 = 262 + 17 y = 26/ - 207\\ +17 310 - 2691 + 17 155 - 56 155 Now put x, y, z in the function, f f (x, y , 2 ) = 2x2 + 3 42 + 2 2 2 2 = 2 61 + 3 - 56 + 2 - 207 155 310 = 3. 219 This is the minimum value of function, f ( my , 2 )1 = 3. 22put a and y in function, fo f ( x , y , 2 ) = 2 2 + 3 42 +2 2 2 = 2 ( - 152 - 11 ) + 3 ( 262 + 17 ) + 2 2 2 - 2 ( 2 25 2 2 + 121 + 3302) + 3( 876 2 2+ 289+ 8842) + 2 22 = 2480 2+ 1109 + 3 312 2 = 2480 2 + 3312 2+ 1109 The min value occurs at 2 = - 6! 2a Then we have , a = 2480, 6 = 3312 - 3312 1656 87:5 419 207 Z = 2 ( 2480) 1240 60 210 2 = - 207 310 m = - 152- 11 21 = + 15/207) 310 - 1 1 Ja = 621 - 1.1 62 2 2 - 61 62\f