Question

i need help with all parts. i will rate.

thank you very much.
f(x,y) = xy on the line 4x+9y=36 using Lagrange Multipliers, the value of is 2 O 2 3 O2

Answer 1

More the objective is to maximize or minimise the function fm,y). So fu,y) is called djective function constraints restricts the points which we should consider we know, Legrange multipliers are used to find the minimum or foras maximum of the function from Lagrange multiplier theorem, Any stationary point of the function that satisfies the equality constraints, the gradient of the function at that foint can be expressed as a linear combination of the gradients of the constraints at the point. If f. (M, g) is objective function & g(ny).org the gillen constraint, Then e fu,y) d g(n,g). Here d is the legrange multipler ie cs ned with CamScanner

Mere, objective feinetion f(wy)=xy and constrainst id untay=36 from Lagrangs multiplier, of de let gcn,y) = untay-3620 » lets find of ve - [Bait ] [ay] I cay) it g (ny) j 2 ay y (181) i tu (iglyj of f = i tu ܠ ܓ y & f ele s elf = Đf y taj lets find og 79=[ 3 nits fys)[untay-35 279= 3 Cantay-36) it J Cantaya3j & Tg GSscanned with CamScanner

>79-6914+) i + (o+copy); 79 ܪ ܢܐ + q j Twhile partially derivating with a y remains as constant as vice versas e kan= knany ; k is constant News, - = of=289 syitni x[uitas » yitni udit ad compare the corresponding elements Juday slad:a y consider the constraint untay=36 u(ad) + q (ud) = 36 364 +36d 3 - 36 > 72 d = 36 da 36 CS Scanned with CamScanner 72

2.11a 2 2) Criven position vector r(t) = {cost, sint, t²). we know, velocity w(t) = f'(t) and acceleration act) = v'ct) so consider r(t) = {cost, sint, te) derivate both sides with t 6 ? s de glo 7 a r(t) Cose, & sint, dt "CAS =<-sint, cost, at> det >> ragain derivate both sides with t. darices - Coff C-Sint) cast "(t) (-cost, -sint, a i UCE) = {-sint, cost, 2t> CScanne a cono <-cost, -sint, ?>

orthogonal of ture vectors a and ħ are Then a. To =0 so, consider VCE) and act). TCE). ä(t) = 0 > (sintitest j+2+$.fcesti-sintit2K => Sint cost - Cost Sint +(2018) -0 tut:0 W t =0 velocity and acceleration are orthogonal only at time to CSScanned with CamScanner

18. 3) Given vector (10, 12,4%. is normal to surface nity² + 24 =18 we know, Gradient of a vector is Normal to the swface The gradient is obtained as follows f(0, y, z) (fx,fy, fą) let f(0, y, z) x² + y²4 qu. partially derivate with a (2²+43 +24_18) -18.20 온 3 on 20002-.to > for » fu 20 partially derivate with y o a a jf by (² + y² + qu. +3ye to el 17 a CSScanned with Camscaler

Partially derivate & with a to & = stad az fz f (x² + y² +2418) cot uz? fz =u23 : orthogonal vector is (fon, fy, fz) 3y?, 423) compare with gillen vector (22,34², uz33 (2x, (10,12 12,4) 34²=12. ;483-4 & 2n = 10 23 > nas ا= 2 clearly from options (5,-2, 1) is the required froint [3(-2)² =12 2:12) Required point is (5,-2, 1) » 12 =12 CSscanned with CamScanner