Question

min / max {5$y2 +2} such that : (a + 2) r² + 3y2 = 6, where a is equal to the product of the two last digits in your ID. For example :

Alpha = 30

Please solve using lagrange method. Find both min and max.

Answer 1

m Let f(xy? = 5xy?+2 and g(x,y) = (x+2) x² + 3y? = 32x2 + 3y26 (:2=0) Let the Lagrange's function be, L2.7, ) = f(x, y) - agraig), 9 being Lagrange's mile, L(, a) = (5«y*+2) – 2( 32x +342-6) At any awitical point de reto, aby so, da so multiplier. mwo NO 01 o- Reg-kerel (= o- the southern 0-9-heteuze ®— -

From ③ we have, y (1ox – 6a) = 0 2 -- 0 3450 or, x Now, yzo and 6 gines, 32x2 - 6 7*= 2 => x = we have two critical points 1 3 o) (410) Now, putting a = 2/3 in ① we have, 5y2 – 642 x 2 = 0. 792-19 22 7 y = 1845 2 - Putting x = 2/3 and y = I 83 in ③ we have 32x 92 + 3x 192 73 - 6 192 25 8640 25 we z 3 tason 7 1= + 1. Put a = 8 in 0 and 6 , we have, x= and y = + 8 3 x 3 1/2 = 74 = + 213 2 putting a r in 9 and we lo que x=-5, jatora

on we h Now, we have another 4 critical points (His) (1 m), 64 ha), 6*1 ) Now, are check the functional value in are total o critical points f(38,0) = 2, f67,0) = 2 f(t. ) = f(4-3) = 5 f(-4,-) = , f6+ *) = i Maximum realue of f(x,y) is H and it is attained af (* ) and (* ) : Minimum value of f(x,y) is, &, and it is attained at (- ):671 nM