Question

If x²+y²=1, then find the maximum value of x²+4xy-y²​

Answer 1

Answer:

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Answer 2

Answer :

√5

Solution :

• Given : x² + y² = 1
• To find : max. value of x² + 4xy - y²

Let x = cos∅ and y = sin∅ .

Then ,

=> x² + y² = cos²∅ + sin²∅

=> x² + y² = 1

Clearly ,

x² + y² = 1 if we assume x = cos∅ and y = sin∅ .

Now ,

x² + 4xy - y²

= cos²∅ + 4cos∅sin∅ - sin²∅

= (cos²∅ - sin²∅) + 2•(2sin∅cos∅)

= cos2∅ + 2sin2∅

Thus ,

x² + 4xy - y² = cos2∅ + 2sin2∅

= 2sin2∅ + 1cos2∅

Also ,

We know that the maximum value of a.sinØ + b.cosØ is √(a² + b²)

Thus ,

The maximum value of 2sin2∅ + 1cos2∅ is √(1² + 2²) = √(1 + 4) = √5

Hence ,

The maximum value of x² + 4xy - y² is √5

Moreover

The minimum value of a.sinØ + b.cosØ is –√(a² + b²)

Thus ,

The minimum value of 2sin2∅ + 1cos2∅ is –√(1² + 2²) = –√(1 + 4) = –√5

Hence ,

The minimum value of x² + 4xy - y² is –√5