C Home Chegg x WA 160 Homework B Section x M Mathway Math Pr

C Home Chegg x WA 160 Homework B Section x M Mathway Math Problem x yl C fi https:// 13086951 www.webassign.ne ty in Linkedin W. Math 160 BroncoJobs B BB P Active Courses FD The vistaIApartment E LU CS internshi s E9 Apps Go RANGERS B B -1 points TanApCalcBr10 8.5.008 My Notes Ask Your Teacher Use the method of Lagrange multipliers to maxim ze the function subject to the given constraint Maximize the function f(x, y) xy subject to the constraint 2x 3y 6 0. at (x, y) -11 points TanApCalcBr10 8.5.010 My Notes Ask Your Teacher Use the method of Lagrange multipliers to minim ze the function subject to the given constraint y 2 x2 subject to the constraint x 2y 5 0. Maximize the function f(x, y) of at (x, y) -11 points TanApCalcBr10 8.5.014 My Notes Ask Your Teacher Use the method of Lagrange multipliers to maxim ze the function subject to the given constraint Maxim the function f(x, y, z) xyz subject to 2x 2y z 84 and x 2 0, y 20, z 0. maximum of at (x, y, z) A 4x 4/29/2016 I\'m Cortana. Ask me anything

Solution

Solution :1)

This method basically says that if we wish to maximize f(x,y) with a constraint of g(x,y) = c, we introduce the variable called the Lagrange multiplier.

Then we introduce a new function dependent on x, y, and . We use the symbol for this new function:
(x,y,) = f(x,y) + *(g(x,y) - c)

For yours we have:
(x,y,) = xy + *(2x + 3y - 6)

The partials of will yield the stationary point if set equal to zero:
/x = y + 2 = 0
/y = x + 3 = 0
/ = 2x + 3y - 6 = 0

Solve the system of equations. We have x = -3 and y = -2. Plug these into the last equation to get:
2(-3) + 3(-2) - 6 = 0
-12 = 6
= -1/2

Then:
x = -3(-1/2) = 3/2
y = -2(-1/2) = 1

Find the z-value associated with this:
f(3/2, 1) = (3/2) * (1) = 3/2

Thus the maximum on the constraint is 3/2 which is attained at (x,y) = (3/2, 1)

Solution : 3)

f(x, y, z) = xyz
g(x,y,z) = 2x + 2y + z - 84
F(x,y,z,) = xyz - (2x + 2y + z - 84)
Fx = yz - 2 = 0 => yz = 2 => = yz / 2
Fy = xz - 2 = 0 => xz = 2 = 2(yz / 2) = yz => x = y
Fz = xy - = 0 => xy = = yz / 2 => x = z/2 => z = 2x
F = - (2x + 2y + z - 84) = 0
2x + 2y + z = 84
2x + 2x + 2x = 84
6x = 84
x = 14
y = 14
z = 28

f(x, y, z) = xyz = 14*14*28 = 5488

Thus the maximum on the constraint is 5488 which is attained at (x,y,z) = (14,14,28) .