Question

21. [-14 Points] DETAILS TANAPCALC10 8.R.037. Use the method of Lagrange multipliers to optimize the function subject to the given constraints. Find the maximum and minimum values of the function f(x, y) = 2x – 3y + 1 subject to the constraint 2x2 + 3y2 – 320 = 0. At what point does the maximum occur? (x, y) = =( What is the value of f(x, y) at this point? f(x, y) = At what point does the minimum occur? (x, y) = What is the value of f(x, y) at this point? f(x, y) = Submit Answer

Answer 1

f(x1) = 2 - 3 + 4 8 ( x, 4) = 3*+ 3* - 310 = 0 -8) «Я - A3 fx = A 4x fy = d gy d.qx - 3 A. 69 d = A = 1 – 3 СУ Як 1 ч = - 1 ото + 3 1 330 + 2 + 2 1 + = 11. 12 2. +x 310 5 7 d = 1 = + 3 +x 31 165 - А- + 1 16

if d = 1 16 g - 1 회 16 2x 1 16 = -8 8 ㅠ 介 if d 1 16 cs 11 = x 21 기기 10 = -8 9 - 8 (-88) Hence we have (8-8), (-8, 3). & 88 - 3x-8 + | = 41 + ④3) f(-8,a) - X-8 - 3 X 8 + 1 - 31

maximum 000X at (19) = 18, 8) Value of f(x,y) at this point f(x,y) = 41. minimum owy at (344) 1-8,8) value of f(x,y) at this paint flwy)= -39.