Question

If 3x+4y+z=5 find min value of 26(x^2+y^+z^2)

Answer 1

Answer:

Step-by-step explanation:

Let the product P  = xy

Since 3x + 4y = 12, y =3−3x4

P = x∗y =3x−34x2

For P to be maximum,dPdx=0

⇒ 3−32x=0 or x=2

Also, d2Pdx2=−32<0, so x =2 is a point of maxima.

When x =2, y = 3−342 =32

Maximum value of P = xy is = 2∗32 =3

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Answer 2

Answer: Minimum value is 25.

Step-by-step explanation:

Minimize : 26 (x² + y² + z²)

Constrain :  3x + 4y + z = 5

Lagrangian function ⇒ L : 26(x² + y² + z²) - λ(3x + 4y + z - 5)

At extrema dL/dx = dL/dy = dL/dz = dL/dλ = 0

$\frac{dL}{dx}$  = 52x - 3λ  

0 = 52x - 3λ

x = 3λ/52

$\frac{dL}{dy}$  = 52y - 4λ  

0 = 52y - 4λ

y = 4λ/52

$\frac{dL}{dz}$  = 52z - λ  

0 = 52z - λ

z = λ/52

dL/dλ = -3x -4y -z + 5

0  = -9λ/52  -16λ/52 -λ/52 + 5

26λ/52 = 5

λ = 10

x = 3λ/52 =  $\frac{30}{52}$

y = 4λ/52 =  $\frac{40}{52}$

z = λ/52 =  $\frac{10}{52}$

26(x² + y² + z²) = $\frac{26}{(52)^{2} } * (30^{2} +40^{2} +10^{2} )$

26(x² + y² + z²)  =  $\frac{26 * 2600}{52 * 52}$

26(x² + y² + z²) = 25

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