Question

(1 point) Use Lagrange multipliers to find the maximum and minimum values of f(x, y, z) = x + 5y + 4z, subject to the constraint x2 + y2 + z2 = 9, if such values exist. maximum = minimum = (For either value, enter DNE if there is no such value.)

Answer 1

The given objective function is,

f = 1 + 5y +42

The given constraint function is,

g=?? + y +9

find the gradients for both the functions,

vf = (1, 5, 4)

V9 = (2:1, 2y, 22)

The Lagrange multiplier eqn is given by,

vt = avg

1 = a(2.c) = I=

5 = a(2y) = y = 2

4= a(22) ► 2=

Now plugin all the values in constraint eqn,

<2 + +22 - 9

$\left ( \frac{1}{2a} \right )^{2}+\left ( \frac{5}{2a} \right )^{2}+\left ( \frac{4}{2a} \right )^{2} = 9$

16 1 25 4a2 + 492 +

1 + 25 + 16 402 = 9

42 49

$\frac{42}{36} = a^{2}$

|-]

A =+ F16

Now x, y, z values become,

1 /6 c= 17

5 /6 y= -1 6-7

4 /6 2%3Dオー

Hence the maximum value is given by,

1 16 5 6 4 6 10 6 16 Imar = 2V 7+2 7+2V 7 = 2V7="17

and the minimum value is given by,

1/6 5 6 4 6 10 6 I min = -21 7-21 7-21 7=27= -17

I hope this answer helps,
Thanks,
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