Question

When x, y and z are real, the minimum value of (2x2 + 2y2 + 5z2 - 2xy - 4yz - 4zx - 4x - 2z + 15) is
(a) 18
(b) 25
(c) 10
(d) 15

Answer 1

Given : 2x² + 2y² + 5z² - 2xy - 4yz  - 4x - 2z + 15  

To Find :  the minimum value

(a) 18

(b) 25

(c) 10

(d) 15

Solution:

2x² + 2y² + 5z² - 2xy - 4yz  - 4x - 2z + 15

= x² + x²  + y² + y² + (2z)² + z² - 2xy - 4yz - 4x - 2z + 15

= x²  - 4x  + x²  + y² - 2xy + y² + (2z)²   - 4yz  + z² - 2z + 15

= ( x - 2)² - 4  + ( x - y)² + (y - 2z)²  + ( z - 1)² - 1  + 15

= ( x - 2)² + (x - y)² + ( y - 2z)²  + ( z - 1)² + 10

x = 2 , y = 2 and z = 1  will give minimum  value

= 10

Learn more:

if x+y+z=0 then the square of the value of (x+y)^2/xy+(y+z)^2/yz+(z+x)

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Find x,y,z using cramers rule, if x-y+z=4, 2x+y-3z=0 and x+y+z=2

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Answer 2

Given :

$\sf \: 2x² + 2y² + 5z² - 2xy - 4yz - 4x - 2z + 15$

To find :

• Minimum value .

Solution :

We know ,

$\rm 2x² + 2y² + 5z² - 2xy - 4yz - 4x - 2z + 15$

$\rm \: = x² + x² + y² + y² + (2z)² + z² - 2xy - 4yz - 4x - 2z + 15$

$\rm \: = x² - 4x + x² + y² - 2xy + y² + (2z)² - 4yz + z²-2z+15$

Let,

$\rm \: = (x - 2)² - 4 + (x - y)² + (y - 2z)² + (z-1)²-1 +15$

$\rm \: = (x - 2)² + (x - y)² + (y-2z)² + (z − 1)² + 10 -</p><p></p><p>x = 2, y = 2 \: and \: z \: = \: 1$

$= 10$

Hence, the minimum value is 10 .